Coordinate Geometry - Straight Line

Find the bisector of the angle in which the origin lies, between the pair of straight lines $\;$ $y - b = \dfrac{2m}{1 - m^2} \left(x - a\right)$ $\;$ and $\;$ $y - b = \dfrac{2m_1}{1 - m_1^2} \left(x - a\right)$.

Equations of the given lines are:

$y - b = \dfrac{2m}{1 - m^2} \left(x - a\right)$ $\;\;\; \cdots \; (1)$

$y - b = \dfrac{2m_1}{1 - m_1^2} \left(x - a\right)$ $\;\;\; \cdots \; (2)$

Equation $(1)$ can be rewritten as

$\left(1 - m^2\right) \left(y - b\right) = 2m \left(x - a\right)$

i.e. $\;$ $\left(1 - m^2\right)y - b + m^2 b = 2mx - 2ma$

i.e. $\;$ $2mx - \left(1 - m^2\right) y - \left(2ma - b + m^2b\right) = 0$ $\;\;\; \cdots \; (1a)$

Equation $(2)$ can be rewritten as

$\left(1 - m_1^2\right) \left(y - b\right) = 2m_1 \left(x - a\right)$

i.e. $\;$ $\left(1 - m_1^2\right)y - b + m_1^2 b = 2m_1x - 2m_1a$

i.e. $\;$ $2m_1x - \left(1 - m_1^2\right) y - \left(2m_1a - b + m_1^2b\right) = 0$ $\;\;\; \cdots \; (2a)$

$\therefore \;$ The equation to the bisector of the angle in which the origin lies is

$\dfrac{2mx - \left(1 - m^2\right) y - \left(2ma - b + m^2b\right)}{\sqrt{\left(2m\right)^2 + \left(1 - m^2\right)^2}}$
$\hspace{2cm}$ $= \dfrac{2m_1x - \left(1 - m_1^2\right) y - \left(2m_1a - b + m_1^2b\right)}{\sqrt{\left(2m_1\right)^2 + \left(1 - m_1^2\right)^2}}$

i.e. $\;$ $\dfrac{2mx - \left(1 - m^2\right) y - \left(2ma - b + m^2b\right)}{\sqrt{4m^2 + 1 - 2m^2 + m^4}}$
$\hspace{2cm}$ $= \dfrac{2m_1x - \left(1 - m_1^2\right) y - \left(2m_1a - b + m_1^2b\right)}{\sqrt{4m_1^2 + 1 - 2m_1^2 + m_1^4}}$

i.e. $\;$ $\dfrac{2mx - \left(1 - m^2\right) y - \left(2ma - b + m^2b\right)}{\sqrt{2m^2 + 1 + m^4}}$
$\hspace{2cm}$ $= \dfrac{2m_1x - \left(1 - m_1^2\right) y - \left(2m_1a - b + m_1^2b\right)}{\sqrt{2m_1^2 + 1 + m_1^4}}$

i.e. $\;$ $\dfrac{2mx - \left(1 - m^2\right) y - \left(2ma - b + m^2b\right)}{\sqrt{\left(1 + m^2\right)^2}}$
$\hspace{2cm}$ $= \dfrac{2m_1x - \left(1 - m_1^2\right) y - \left(2m_1a - b + m_1^2b\right)}{\sqrt{\left(1 + m_1^2\right)^2}}$

i.e. $\;$ $\dfrac{2mx - \left(1 - m^2\right) y - \left(2ma - b + m^2b\right)}{1 + m^2}$
$\hspace{2cm}$ $= \dfrac{2m_1x - \left(1 - m_1^2\right) y - \left(2m_1a - b + m_1^2b\right)}{1 + m_1^2}$

i.e. $\;$ $2m \left(1 + m_1^2\right) x - \left(1 - m^2\right) \left(1 + m_1^2\right)y - \left(2ma - b + m^2b\right) \left(1 + m_1^2\right)$
$\hspace{1cm}$ $= 2m_1 \left(1 + m^2\right) x - \left(1 - m_1^2\right) \left(1 + m^2\right)y - \left(2m_1a - b + m_1^2b\right) \left(1 + m^2\right)$

i.e. $\;$ $2\left(m + m m_1^2 - m_1 - m_1 m^2\right)x - \left(1 + m_1^2 - m^2 - m^2 m_1^2 -1 - m^2 + m_1^2 + m^2 m_1^2\right)y$
$\hspace{1cm}$ $= 2ma + 2mm_1^2 a - b - b m_1^2 + bm^2 + bm^2 m_1^2$
$\hspace{1.5cm}$ $-2m_1a - 2m_1 m^2 a + b + bm^2 - b m_1^2 - b m^2 m_1^2$

i.e. $\;$ $x \left(m + m m_1^2 - m_1 - m_1 m^2\right) - y \left(m_1^2 - m^2\right)$
$\hspace{1cm}$ $= ma + m m_1^2 a - b m_1^2 + bm^2 - m_1 a - m_1 m^2 a$

i.e. $\;$ $x \left(m + m m_1^2 - m_1 - m_1 m^2\right) - y \left(m_1^2 - m^2\right)$
$\hspace{1cm}$ $= a \left(m + m m_1^2 - m_1 - m_1 m^2\right) - b \left(m_1^2 - m^2\right)$

i.e. $\;$ $\left(x - a\right) \left(m + m m_1^2 - m_1 - m_1 m^2\right) - \left(y - b\right) \left(m_1^2 - m^2\right) = 0$

i.e. $\;$ $\left(x - a\right) \left[\left(m - m_1\right) + mm_1 \left(m_1 - m\right)\right] - \left(y - b\right) \left(m_1 - m\right) \left(m_1 + m\right) = 0$

i.e. $\;$ $\left(x - a\right) \left[- \left(m_1 - m\right) + mm_1 \left(m_1 - m\right)\right] - \left(y - b\right) \left(m_1 - m\right) \left(m_1 + m\right) = 0$

i.e. $\;$ $\left(x - a\right) \left(mm_1 - 1\right) \left(m_1 - m\right) - \left(y - b\right) \left(m_1 - m\right) \left(m_1 + m\right) = 0$

i.e. $\;$ $\left(x - a\right) \left(mm_1 - 1\right) - \left(y - b\right) \left(m_1 + m\right) = 0$

Coordinate Geometry - Straight Line

Find the bisector of the angles between the pair of straight lines $4x + 3y = 3$ and $5x - 12y = 8$, placing first the bisector of that angle in which the origin lies.

Equations of the given lines are: $\;\;$ $4x + 3y = 3$, $\;\;\;$ $5x - 12y = 8$

Writing the equations so that their constant terms are negative, they are

$4x + 3y -3 = 0$, $\;\;\;$ $5x - 12y - 8 = 0$

$\therefore \;$ The equation to the bisector of the angle in which the origin lies is

$\dfrac{4x + 3y - 3}{\sqrt{4^2 + 3^2}} = \dfrac{5x - 12y - 8}{\sqrt{5^2 + 12^2}}$

i.e. $\;$ $\dfrac{4x + 3y - 3}{5} = \dfrac{5x - 12y - 8}{13}$

i.e. $\;$ $13 \left(4x + 3y - 3\right) = 5 \left(5x - 12y - 8\right)$

i.e. $\;$ $52 x + 39 y - 39 = 25 x - 60 y - 40$

i.e. $\;$ $27 x + 99y + 1 = 0$

The equation to the other bisector is

$\dfrac{4x + 3y - 3}{\sqrt{4^2 + 3^2}} = \dfrac{- \left(5x - 12y - 8\right)}{\sqrt{5^2 + 12^2}}$

i.e. $\;$ $\dfrac{4x + 3y - 3}{5} = \dfrac{- \left(5x - 12y - 8\right)}{13}$

i.e. $\;$ $13 \left(4x + 3y - 3\right) = -5 \left(5x - 12y - 8\right)$

i.e. $\;$ $52 x + 39 y - 39 = -25 x + 60 y + 40$

i.e. $\;$ $77 x - 21y - 79 = 0$

Coordinate Geometry - Straight Line

Find whether the points $\left(2, -3\right)$ and $\left(3, -2\right)$ are on the same side or on opposite sides of the line $5x + 6y = 4$.

Equation of given line: $\;\;$ $5x + 6y = 4$ $\;$ i.e. $\;$ $5x + 6y -4 = 0$ $\;\;\; \cdots \; (1)$

Given points: $\;\;$ $A \left(x_1, y_1\right) = \left(2, -3\right)$; $\;$ $B \left(x_2, y_2\right) = \left(3, -2\right)$

Substituting $\;$ $x = 2, \; y = -3$ $\;$ in $\;$ $5x + 6y - 4$, $\;$ it becomes

$5 \times 2 + 6 \times \left(-3\right) - 4 = -12$ $\;$ i.e. $\;$ it is negative.

Substituting $\;$ $x = 3, \; y = -2$ $\;$ in $\;$ $5x + 6y - 4$, $\;$ it becomes

$5 \times 3 + 6 \times \left(-2\right) - 4 = -1$ $\;$ i.e. $\;$ it is negative.

Hence the points lie on the same side of the given line.

Coordinate Geometry - Straight Line

Find the equation of the straight line parallel to the lines $2x + 3y = 5$ and $2x + 3y + 4 = 0$, which divides the distance between them internally in the ratio $3 : 5$.

The given lines are: $\;$ $2x + 3y = 5$ $\;\;\; \cdots \; (1)$; $\;$ $2x + 3y + 4 = 0$ $\;\;\; \cdots \; (2)$

Putting $y = 0$ in equation $(1)$ gives $\;\;$ $x = \dfrac{5}{2}$

$\implies$ Equation $(1)$ cuts the $X$ axis at the point $A \left(\dfrac{5}{2}, 0\right)$

Putting $y = 0$ in equation $(2)$ gives $\;\;$ $x = -2$

$\implies$ Equation $(2)$ cuts the $X$ axis at the point $B \left(-2, 0\right)$

Let $P \left(x_1, 0\right)$ be a point on the required line which cuts the line segment $AB$ internally in the ratio $3 : 5$

Then by section formula for internal division,

$x_1 = \dfrac{3 \times \left(-2\right) + 5 \times \dfrac{5}{2}}{3 + 5} = \dfrac{-6 + \dfrac{25}{2}}{8} = \dfrac{13}{16}$

i.e. $\;$ $P \left(x_1, 0\right) = \left(\dfrac{13}{16}\right)$

Slope of line given by equation $(1)$ is $m = \dfrac{-2}{3}$

Since the required line is parallel to lines given by equations $(1)$ and $(2)$,

slope of required line $= m = \dfrac{-2}{3}$

and the required line passes through the point $P$.

$\therefore \;$ The equation of the required line is

$y - 0 = \dfrac{-2}{3} \left(x - \dfrac{13}{16}\right)$

i.e. $\;$ $3y = -2x + \dfrac{13}{8}$

i.e. $\;$ $16 x + 24 y - 13 = 0$

Coordinate Geometry - Straight Line

Find the distance between the parallel lines $5x - 3y + 4 = 0$ and $10x - 6y - 9 = 0$

Equations of the given parallel lines are

$5x - 3y + 4 = 0$ $\;\;\; \cdots \; (1)$; $\;$ $10x - 6y - 9 = 0$ $\;\;\; \cdots \; (2)$

The perpendicular distance from $\left(0,0\right)$ to equation $(1)$ is

$= \left|\dfrac{4}{\sqrt{5^2 + \left(-3\right)^2}}\right| = \dfrac{4}{\sqrt{34}}$

The perpendicular distance from $\left(0,0\right)$ to equation $(2)$ is

$= \left|\dfrac{-9}{\sqrt{10^2 + \left(-6\right)^2}}\right| = \dfrac{9}{\sqrt{136}} = \dfrac{9}{2 \sqrt{34}}$

Substituting $y =0$ in equation $(1)$ gives $\;\;$ $x = \dfrac{-4}{5}$

$\implies$ Equation $(1)$ cuts off negative intercept on the $X$ axis.

Substituting $y =0$ in equation $(2)$ gives $\;\;$ $x = \dfrac{9}{10}$

$\implies$ Equation $(2)$ cuts off positive intercept on the $X$ axis.

$\therefore \;$ The origin lies between the lines given by equations $(1)$ and $(2)$.

$\therefore \;$ The distance between the parallel lines is

$= \dfrac{4}{\sqrt{34}} + \dfrac{9}{2 \sqrt{34}} = \dfrac{17}{2 \sqrt{34}} = \dfrac{34}{4 \sqrt{34}} = \dfrac{\sqrt{34}}{4}$

Coordinate Geometry - Straight Line

Find the equations to the lines through the point $\left(1, 1\right)$ which are $3$ units distant from the point $\left(-2, 3\right)$.

Let the equation of the required line be $\;$ $Ax + By + C = 0$ $\;\;\; \cdots \; (1)$

Distance of the required line from the point $\left(x_1, y_1\right) = \left(-2, 3\right)$ $\;$ is $\;$ $= 3$ units

i.e. $\;$ $\dfrac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}} = 3$

i.e. $\;$ $-2 A + 3 B + C = 3 \sqrt{A^2 + B^2}$ $\;\;\; \cdots \; (2)$

Since the required equation passes through the point $\left(1,1\right)$, we have

$A + B + C = 0$ $\implies$ $C = - A - B$ $\;\;\; \cdots \; (3)$

Substituting the value of $C$ in equation $(2)$ gives

$- 2A + 3B - A - B = 3 \sqrt{A^2 + B^2}$

i.e. $\;$ $-3A + 2B = 3 \sqrt{A^2 + B^2}$

i.e. $\;$ $9A^2 + 4B^2 -12AB = 9A^2 + 9B^2$

i.e. $\;$ $5B^2 = - 12 AB$

i.e. $\;$ $B \left(5 B + 12 A\right) = 0$ $\;\;\; \cdots \; (4)$

i.e. $\;$ $B = 0$ $\;\;\;$ OR $\;\;\;$ $5B + 12A = 0$

When $\;$ $5B + 12A = 0$ $\implies$ $A = \dfrac{-5}{12} B$

Substituting the value of $A$ in equation $(3)$ gives

$C = \dfrac{5}{12} B - B = \dfrac{-7}{12} B$

Substituting the values of $A$ and $C$ in equation $(1)$ gives

$\dfrac{-5}{12} B x + B y - \dfrac{7}{12} B = 0$

i.e. $\;$ $-5 x + 12 y - 7 = 0$

$\therefore \;$ The equation of the required line is $\;\;$ $5x - 12 y + 7 = 0$

When $\;$ $B = 0$, $\;$ we have from equation $(3)$, $\;\;\;$ $C = -A$

Substituting $C = -A$ and $B = 0$ in equation $(1)$ gives

$Ax - A = 0$

$\therefore \;$ The equation of the required line is $\;\;$ $x - 1 = 0$

Coordinate Geometry - Straight Line

Find the length of the altitudes of the triangle whose sides are the lines $3x - 4y = 5$, $4x + 3y = 5$ and $x + y = 1$.

The sides of the triangle are

$3x - 4y = 5$ $\;\;\; \cdots \; (1)$, $\;$ $4x + 3y = 5$ $\;\;\; \cdots \; (2)$, $\;$ $x + y = 1$ $\;\;\; \cdots \; (3)$

Let equations $(1)$ and $(2)$ intersect at point $A$.

Let equations $(2)$ and $(3)$ intersect at point $B$.

Let equations $(3)$ and $(1)$ intersect at point $C$.

Solving equations $(1)$ and $(2)$ simultaneously gives the point of intersection as $\;$ $A \left(x_1, y_1\right) = \left(\dfrac{7}{5}, \dfrac{-1}{5}\right)$

Solving equations $(2)$ and $(3)$ simultaneously gives the point of intersection as $\;$ $B \left(x_2, y_2\right) = \left(2, -1\right)$

Solving equations $(3)$ and $(1)$ simultaneously gives the point of intersection as $\;$ $C \left(x_3, y_3\right) = \left(\dfrac{9}{7}, \dfrac{-2}{7}\right)$

Let $AD$ be the altitude drawn from point $A$ to line given by equation $(3)$.

Equation $(3)$ can be written as $\;$ $x + y - 1 = 0$

$\therefore \;$ Length of $AD = \dfrac{1 \times \dfrac{7}{5} + 1 \times \left(\dfrac{-1}{5}\right) - 1}{\sqrt{1^2 + 1^2}} = \dfrac{1}{5 \sqrt{2}}$

Let $BE$ be the altitude drawn from point $B$ to line given by equation $(1)$.

Equation $(1)$ can be written as $\;$ $3x - 4y - 5 = 0$

$\therefore \;$ Length of $BE = \dfrac{3 \times 2 - 4 \times \left(-1\right) - 5}{\sqrt{3^2 + \left(-4\right)^2}} = \dfrac{5}{5} = 1$

Let $CF$ be the altitude drawn from point $C$ to line given by equation $(2)$.

Equation $(2)$ can be written as $\;$ $4x + 3y - 5 = 0$

$\therefore \;$ Length of $CF = \dfrac{4 \times \dfrac{9}{7} + 3 \times \left(\dfrac{-2}{7}\right) - 5}{\sqrt{4^2 + 3^2}} = \dfrac{1}{5}$

Coordinate Geometry - Straight Line

Find the perpendicular distance from the origin to the line joining the two points $\left(a \cos \alpha, a \sin \alpha\right)$ and $\left(a \cos \beta, a \sin \beta\right)$.

Given points are: $\;$ $\left(x_1, y_1\right) = \left(a \cos \alpha, a \sin \alpha\right)$; $\;$ $\left(x_2, y_2\right) = \left(a \cos \beta, a \sin \beta\right)$

Slope of the line through the points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is

$m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{a \sin \beta - a \sin \alpha}{a \cos \beta - a \cos \alpha} = \dfrac{\sin \beta - \sin \alpha}{\cos \beta - \cos \alpha}$

i.e. $\;$ $m = \dfrac{2 \cos \left(\dfrac{\beta + \alpha}{2}\right) \sin \left(\dfrac{\beta - \alpha}{2}\right)}{-2 \sin \left(\dfrac{\beta + \alpha}{2}\right) \sin \left(\dfrac{\beta - \alpha}{2}\right)} = \dfrac{- \cos \left(\dfrac{\alpha + \beta}{2}\right)}{\sin \left(\dfrac{\alpha + \beta}{2}\right)}$

$\therefore \;$ Equation of the line through the points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is of the form

$y - y_1 = m \left(x - x_1\right)$

i.e. $\;$ $y - a \sin \alpha = \left[\dfrac{- \cos \left(\dfrac{\alpha + \beta}{2}\right)}{\sin \left(\dfrac{\alpha + \beta}{2}\right)}\right] \left(x - a \cos \alpha\right)$

i.e. $\;$ $y \sin \left(\dfrac{\alpha + \beta}{2}\right) - a \sin \alpha \sin \left(\dfrac{\alpha + \beta}{2}\right) = - x \cos \left(\dfrac{\alpha + \beta}{2}\right) + a \cos \alpha \cos \left(\dfrac{\alpha + \beta}{2}\right)$

i.e. $\;$ $x \cos \left(\dfrac{\alpha + \beta}{2}\right) + y \sin \left(\dfrac{\alpha + \beta}{2}\right) - a \left[\sin \alpha \sin \left(\dfrac{\alpha + \beta}{2}\right) + \cos \alpha \cos \left(\dfrac{\alpha + \beta}{2}\right)\right] = 0$

i.e. $\;$ $x \cos \left(\dfrac{\alpha + \beta}{2}\right) + y \sin \left(\dfrac{\alpha + \beta}{2}\right) - a \cos \left(\alpha - \dfrac{\alpha + \beta}{2}\right) = 0$

i.e. $\;$ $x \cos \left(\dfrac{\alpha + \beta}{2}\right) + y \sin \left(\dfrac{\alpha + \beta}{2}\right) - a \cos \left(\dfrac{\alpha - \beta}{2}\right) = 0$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of a straight line $Ax + By + C = 0$ gives

$A = \cos \left(\dfrac{\alpha + \beta}{2}\right)$, $\;$ $B = \sin \left(\dfrac{\alpha + \beta}{2}\right)$, $\;$ $C = - a \cos \left(\dfrac{\alpha - \beta}{2}\right)$

Given point: $\left(x_3, y_3\right) = \left(0, 0\right)$

Perpendicular distance from the point $\left(x_3, y_3\right)$ to the line $Ax + By + C = 0$ is $= \dfrac{Ax_3 + By_3 + C}{\sqrt{A^2 + B^2}}$

$\therefore \;$ Required perpendicular distance is

$= \left|\dfrac{- a \cos \left(\dfrac{\alpha - \beta}{2}\right)}{\sqrt{\cos^2 \left(\dfrac{\alpha + \beta}{2}\right) + \sin^2 \left(\dfrac{\alpha + \beta}{2}\right)}}\right| = \left|- a \cos \left(\dfrac{\alpha - \beta}{2}\right)\right| = a \cos \left(\dfrac{\alpha - \beta}{2}\right)$

Coordinate Geometry - Straight Line

Find the perpendicular distance from the point $\left(3, 4\right)$ to the straight line $3x - 4y + 10 = 0$

Equation of given line is $\;$ $3x - 4y + 10 = 0$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of a straight line $Ax + By + C = 0$ gives

$A = 3$, $\;$ $B = -4$, $\;$ $C = 10$

Given point: $\left(x_1, y_1\right) = \left(3, 4\right)$

Perpendicular distance from the point $\left(x_1, y_1\right)$ to the line $Ax + By + C = 0$ is $= \dfrac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}}$

$\therefore \;$ Required perpendicular distance is

$= \dfrac{3 \times 3 - 4 \times 4 + 10}{\sqrt{3^2 + \left(-4\right)^2}} = \dfrac{9 - 16 + 10}{\sqrt{25}} = \dfrac{3}{5}$

Coordinate Geometry - Straight Line

For what value of $a$ are the lines $3x + y + 2 = 0$, $2x - y+ 3 = 0$ and $x + ay - 3 = 0$ are concurrent.

Equations of the given lines are

$3x + y + 2 = 0$ $\;\;\; \cdots \; (1)$; $\;$ $2x - y + 3 = 0$ $\;\;\; \cdots \; (2)$; $\;$ $x + ay - 3 = 0$ $\;\;\; \cdots \; (3)$

Solving equations $(1)$ and $(2)$ simultaneously we get

$5x + 5 = 0$ $\implies$ $x = -1$

and $y = 2 \times \left(-1\right) + 3 = 1$

$\therefore \;$ The point of intersection of equations $(1)$ and $(2)$ is $\left(x, y\right) = \left(-1, 1\right)$

$\because \;$ Equations $(1)$, $(2)$ and $(3)$ are concurrent, the point $\left(x,y\right)$ also satisfies equation $(3)$.

Substituting $\left(x,y\right)$ in equation $(3)$, we get

$-1 + a -3 = 0$ $\implies$ $a = 4$

Coordinate Geometry - Straight Line

Find the equation to the straight line passing through the intersection of the lines $2x - 5y + 1 = 0$ and $x + 3y - 4 = 0$ and cutting off equal intercepts from the axes.

Equation of a line passing through the intersection of the lines

$2x - 5y + 1 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x + 3y - 4 = 0$ $\;\;\; \cdots \; (2)$ is

$\left(2x - 5y + 1\right) + \lambda \left(x + 3y - 4\right) = 0$ $\;\;\; \cdots \; (3)$ $\;$ where $\lambda$ is a scalar.

i.e. $\;$ $\left(2 + \lambda\right)x + \left(-5 + 3 \lambda\right)y + \left(1 - 4 \lambda\right) = 0$ $\;\;\; \cdots \; (4)$

i.e. $\;$ $\left(2 + \lambda\right)x + \left(-5 + 3 \lambda\right)y = \left(4 \lambda - 1\right)$

i.e. $\;$ $\dfrac{\left(2 + \lambda\right) x}{4 \lambda -1} + \dfrac{\left(3 \lambda - 5\right) y}{4 \lambda - 1} = 1$

i.e. $\;$ $\dfrac{x}{\dfrac{4\lambda - 1}{2 + \lambda}} + \dfrac{y}{\dfrac{4 \lambda - 1}{3 \lambda - 5}} = 1$

$\therefore \;$ $X$ intercept of the required line $= a = \dfrac{4 \lambda - 1}{2 + \lambda}$

$Y$ intercept of the required line $= b = \dfrac{4 \lambda - 1}{3 \lambda - 5}$

Given: $\;$ $a = b$

i.e. $\;$ $\dfrac{4 \lambda - 1}{2 + \lambda} = \dfrac{4 \lambda - 1}{3 \lambda - 5}$

i.e. $\;$ $3 \lambda - 5 = 2 + \lambda$

i.e. $\;$ $2 \lambda = 7$ $\implies$ $\lambda = \dfrac{7}{2}$

Substituting the value of $\lambda$ in equation $(4)$ gives the required equation of line as

$\dfrac{11}{2} x + \dfrac{11}{2} y - 13 = 0$

i.e. $\;$ $11 x + 11y - 26 = 0$

Coordinate Geometry - Straight Line

Find the equation to the straight line passing through the intersection of the lines $x - 2y -3 = 0$ and $x + 3y - 6 = 0$ and parallel to the line $3x + 4y = 7$

Equation of a line passing through the intersection of the lines

$x - 2y -3 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x + 3y - 6 = 0$ $\;\;\; \cdots \; (2)$ is

$\left(x - 2y -3\right) + \lambda \left(x + 3y - 6\right) = 0$ $\;\;\; \cdots \; (3)$ $\;$ where $\lambda$ is a scalar.

i.e. $\;$ $\left(1 + \lambda\right)x + \left(-2 + 3 \lambda\right)y - \left(3 + 6 \lambda\right) = 0$ $\;\;\; \cdots \; (4)$

i.e. $\;$ $y = \left(\dfrac{1 + \lambda}{2 - 3 \lambda}\right) x + \dfrac{3 + 6 \lambda}{-2 + 3 \lambda}$

$\therefore \;$ Slope of equation $(3)$ is $= m_1 = \dfrac{1 + \lambda}{2 - 3 \lambda}$

Equation of given line is: $\;$ $3x + 4y = 7$

i.e. $\;$ $y = \dfrac{-3}{4} x + \dfrac{7}{4}$

$\therefore \;$ Slope of given line $= m_2 = \dfrac{-3}{4}$

Since the required line is parallel to the given line, $\;$ $m_1 = m_2$

i.e. $\;$ $\dfrac{1 + \lambda}{2 - 3 \lambda} = \dfrac{-3}{4}$

i.e. $\;$ $4 + 4 \lambda = -6 + 9 \lambda$

i.e. $\;$ $5 \lambda = 10$ $\implies$ $\lambda = 2$

Substituting the value of $\lambda$ in equation $(4)$ gives the equation of the required line as

$3x + 4y - 15 = 0$

Coordinate Geometry - Straight Line

Find the equation to the straight line passing through the point $\left(3,2\right)$ and the intersection of the lines $2x + 3y = 1$ and $3x - 4y = 6$.

Comparing $\;$ $2x + 3y = 1$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $3x - 4y = 6$ $\;\;\; \cdots \; (2)$

with the standard equations $\;$ $a_1x + b_1 y + c_1 = 0$ $\;$ and $\;$ $a_2 x + b_2 y + c_2 = 0$ respectively gives

$a_1 = 2$, $\;$ $b_1 = 3$, $\;$ $c_1 = -1$, $\;$ $a_2 = 3$, $\;$ $b_2 = -4$, $\;$ $c_2 = -6$

The coordinates point of intersection of equations $(1)$ and $(2)$ are

$x = \dfrac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}$ $\;$ and $\;$ $y = \dfrac{c_1 a_2 - c_2 a_1}{a_1 b_2 - a_2 b_1}$

i.e. $\;$ $x = \dfrac{3 \times \left(-6\right) + 4 \times \left(-1\right)}{2 \times \left(-4\right) - 3 \times 3}$ $\;$ and $\;$ $y = \dfrac{-1 \times 3 + 6 \times 2}{2 \times \left(-4\right) - 3 \times 3}$

i.e. $\;$ $x = \dfrac{-22}{-17}$, $\;$ $y = \dfrac{9}{-17}$

i.e. $\;$ The point of intersection is $\;$ $= \left(x, y\right) = \left(\dfrac{22}{17},\dfrac{-9}{17}\right)$

The required equation passes through the points $\left(x_1, y_1\right) = \left(3, 2\right)$ and $\left(x_2, y_2\right) = \left(\dfrac{22}{17},\dfrac{-9}{17}\right)$

Slope of the required line $= m = \dfrac{y_2 - y_1}{x_2 - x_1}$

i.e. $\;$ $m = \dfrac{\dfrac{-9}{17} - 2}{\dfrac{22}{17} - 3} = \dfrac{43}{29}$

Equation of the required line is of the form $\;$ $y - y_1 = m \left(x - x_1\right)$

i.e. $\;$ $y - 2 = \dfrac{43}{29} \left(x - 3\right)$

i.e. $\;$ $29 y - 58 = 43 x - 129$

i.e. $\;$ $43 x - 29y - 71 = 0$

Coordinate Geometry - Straight Line

Find the coordinates of the point of intersection of the lines $2x - 3y = 1$ and $5y - x = 3$, and find the acute angle at which they cut each other.

Comparing $\;$ $2x - 3y = 1$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $5y - x = 3$ $\;\;\; \cdots \; (2)$

with the standard equations $\;$ $a_1x + b_1 y + c_1 = 0$ $\;$ and $\;$ $a_2 x + b_2 y + c_2 = 0$ respectively gives

$a_1 = 2$, $\;$ $b_1 = -3$, $\;$ $c_1 = -1$, $\;$ $a_2 = 1$, $\;$ $b_2 = -5$, $\;$ $c_2 = 3$

The coordinates point of intersection of equations $(1)$ and $(2)$ are

$x = \dfrac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}$ $\;$ and $\;$ $y = \dfrac{c_1 a_2 - c_2 a_1}{a_1 b_2 - a_2 b_1}$

i.e. $\;$ $x = \dfrac{-3 \times 3 + 5 \times \left(-1\right)}{2 \times \left(-5\right) - 1 \times \left(-3\right)}$ $\;$ and $\;$ $y = \dfrac{-1 \times 1 - 3 \times 2}{2 \times \left(-5\right) - 1 \times \left(-3\right)}$

i.e. $\;$ $x = \dfrac{-14}{-7}$, $\;$ $y = \dfrac{-7}{-7}$

i.e. $\;$ The point of intersection is $\;$ $= \left(x, y\right) = \left(2,1\right)$

Equation $(1)$ can be written as $\;$ $y = \dfrac{2}{3}x - \dfrac{1}{3}$ $\;\;\; \cdots \; (3)$

Equation $(2)$ can be written as $\;$ $y = \dfrac{1}{5}x + \dfrac{3}{5}$ $\;\;\; \cdots \; (4)$

Slope of equation $(3)$ is $= m_1 = \dfrac{2}{3}$

Slope of equation $(4)$ is $= m_2 = \dfrac{1}{5}$

Let $\theta$ be the angle between the lines given by equations $(1)$ and $(2)$.

Then $\;$ $\theta = \tan^{-1} \left(\dfrac{m_1 - m_2}{1 + m_1 m_2}\right)$

i.e. $\;$ $\theta = \tan^{-1} \left(\dfrac{\dfrac{2}{3} - \dfrac{1}{5}}{1 + \dfrac{2}{3} \times \dfrac{1}{5}}\right)$

i.e. $\;$ $\theta = \tan^{-1} \left(\dfrac{7}{17}\right)$

Coordinate Geometry - Straight Line

Find the equation of the line passing through $\left(-6, 10\right)$ and perpendicular to the line $7x + 8y = 1$

Equation of given line $\;$ $7x + 8y = 1$

i.e. $\;$ $y = \dfrac{-7}{8} x + \dfrac{1}{8}$

$\therefore \;$ Slope of the given line $= m = \dfrac{-7}{8}$

Since the required line is perpendicular to the given line,

slope of the required line $= \dfrac{-1}{m} = \dfrac{8}{7}$

The required line passes through the point $\left(x_1, y_1\right) = \left(-6, 10\right)$

$\therefore \;$ The equation of the required line is of the form: $\;$ $y - y_1 = m \left(x - x_1\right)$

i.e. $\;$ $y - 10 = \dfrac{8}{7} \left(x + 6\right)$

i.e. $\;$ $7y - 70 = 8x + 48$

i.e. $\;$ $8x - 7y + 118 = 0$

Coordinate Geometry - Straight Line

Find the equation of the line parallel to $\;$ $2x + 5y = 7$ $\;$ and making an intercept of $2$ on the $Y$ axis.

Equation of given line $\;$ $2x + 5y = 7$

i.e. $\;$ $y = \dfrac{-2}{5} x + \dfrac{7}{5}$

$\therefore \;$ Slope of the given line $= m = \dfrac{-2}{5}$

Since the required line is parallel to the given line,

slope of the required line $= m = \dfrac{-2}{5}$

$Y$ intercept of the required line $= c = 2$

$\therefore \;$ The equation of the required line is of the form: $\;$ $y = mx + c$

i.e. $\;$ $y = \dfrac{-2}{5}x + 2$

i.e. $\;$ $2x + 5y - 10 = 0$

Coordinate Geometry - Straight Line

Find the equation of the line passing through $\left(2, 3\right)$ and parallel to the line $5x + 4y = 6$

Equation of given line $\;$ $5x + 4y = 6$

i.e. $\;$ $y = \dfrac{-5}{4} x + \dfrac{6}{4}$

$\therefore \;$ Slope of the given line $= m = \dfrac{-5}{4}$

Since the required line is parallel to the given line,

slope of the required line $= m = \dfrac{-5}{4}$

The required line passes through the point $\left(x_1, y_1\right) = \left(2, 3\right)$

$\therefore \;$ The equation of the required line is of the form: $\;$ $y - y_1 = m \left(x - x_1\right)$

i.e. $\;$ $y - 3 = \dfrac{-5}{4} \left(x - 2\right)$

i.e. $\;$ $4y - 12 = -5x + 10$

i.e. $\;$ $5x + 4y - 22 = 0$

Coordinate Geometry - Straight Line

Find the equations to the straight lines which pass through the origin and are inclined at $75^\circ$ to the straight line $x \left(1 - \sqrt{3}\right) + y \left(1 + \sqrt{3}\right) + c = 0$

Let the given point $= P \left(x_1, y_1\right) = \left(0,0\right)$

Equation of given line: $\;$ $x \left(1 - \sqrt{3}\right) + y \left(1 + \sqrt{3}\right) + c = 0$

i.e. $\;$ $y = \left(\dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}\right) x - \dfrac{c}{\sqrt{3} + 1}$

$\therefore \;$ Slope of given line $= m = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}$

i.e. $\;$ $m = \dfrac{\left(\sqrt{3} - 1\right)^2}{\left(\sqrt{3} + 1\right) \left(\sqrt{3} - 1\right)} = \dfrac{3 + 1 - 2 \sqrt{3}}{3 - 1} = 2 - \sqrt{3}$

Let the angle between the given line and the required lines $= \alpha = 75^\circ$

Then the equations of the required two lines are

$y - y_1 = \left(\dfrac{m - \tan \alpha}{1 + m \tan \alpha}\right) \left(x - x_1\right)$ $\;$ and $\;$ $y - y_1 = \left(\dfrac{m + \tan \alpha}{1 - m \tan \alpha}\right) \left(x - x_1\right)$

$\therefore \;$ The equations of the two lines are

$y - 0 = \dfrac{2 - \sqrt{3} - \tan 75^\circ}{1 + \left(2 - \sqrt{3}\right) \times \tan 75^\circ} \left(x - 0\right)$ $\;$ and $\;$ $y - 0 = \dfrac{2 - \sqrt{3} + \tan 75^\circ}{1 - \left(2 - \sqrt{3}\right) \times \tan 75^\circ} \left(x - 0\right)$

i.e. $\;$ $y = \left[\dfrac{2 - \sqrt{3} - \left(2 + \sqrt{3}\right)}{1 + \left(2 - \sqrt{3}\right) \times \left(2 + \sqrt{3}\right)}\right] x$ $\;$ and $\;$ $y = \left[\dfrac{2 - \sqrt{3} + 2 + \sqrt{3}}{1 - \left(2 - \sqrt{3}\right) \times \left(2 + \sqrt{3}\right)}\right] x$

i.e. $\;$ $y = \left(\dfrac{-2 \sqrt{3}}{1 + 4 - 3}\right) x$ $\;$ and $\;$ $y = \left(\dfrac{4}{1 - 4 + 3}\right) x$

i.e. $\;$ $y = -\sqrt{3} x$ $\;$ and $\;$ $x = \left(\dfrac{0}{4}\right) y$

i.e. $\;$ $\sqrt{3} x + y = 0$ $\;$ and $\;$ $x = 0$

Coordinate Geometry - Straight Line

Find the equations to the two lines through $\left(3, 2\right)$ making an angle of $30^\circ$ with the line $x + 2y = 2$.
Find the area of the triangle enclosed by these two lines and the $X$ axis.

Let the given point $= P \left(x_1, y_1\right) = \left(3,2\right)$

Equation of given line: $\;$ $x + 2y = 2$

i.e. $\;$ $y = \dfrac{-1}{2}x + 1$

$\therefore \;$ Slope of given line $= m = \dfrac{-1}{2}$

Let the angle between the given line and the required lines $= \alpha = 30^\circ$

Then the equations of the required two lines are

$y - y_1 = \dfrac{m - \tan \alpha}{1 + m \tan \alpha} \left(x - x_1\right)$

and $\;$ $y - y_1 = \dfrac{m + \tan \alpha}{1 - m \tan \alpha} \left(x - x_1\right)$

$\therefore \;$ The equations of the two lines are

$y - 2 = \dfrac{\dfrac{-1}{2} - \tan 30^\circ}{1 - \dfrac{1}{2} \times \tan 30^\circ} \left(x - 3\right)$

and $\;$ $y - 2 = \dfrac{\dfrac{-1}{2} + \tan 30^\circ}{1 + \dfrac{1}{2} \times \tan 30^\circ} \left(x - 3\right)$

i.e. $\;$ $y - 2 = \left(\dfrac{\dfrac{-1}{2} - \dfrac{1}{\sqrt{3}}}{1 - \dfrac{1}{2 \sqrt{3}}}\right) \left(x - 3\right)$

and $\;$ $y - 2 = \left(\dfrac{\dfrac{-1}{2} + \dfrac{1}{\sqrt{3}}}{1 + \dfrac{1}{2 \sqrt{3}}}\right) \left(x - 3\right)$

i.e. $\;$ $y - 2 = \left(\dfrac{-\sqrt{3} - 2}{2 \sqrt{3} - 1}\right) \left(x - 3\right)$

and $\;$ $y - 2 = \left(\dfrac{-\sqrt{3} + 2}{2 \sqrt{3} + 1}\right) \left(x - 3\right)$

i.e. $\;$ $y \left(2 \sqrt{3} - 1\right) - 2 \left(2 \sqrt{3} - 1\right) = - \left(\sqrt{3} + 2\right)x + 3 \left(\sqrt{3} + 2\right)$

and $\;$ $y \left(2 \sqrt{3} + 1\right) - 2 \left(2 \sqrt{3} + 1\right) = \left(-\sqrt{3} + 2\right)x - 3 \left(-\sqrt{3} + 2\right)$

i.e. $\;$ $\left(\sqrt{3} + 2\right)x + \left(2 \sqrt{3} -1\right) y = 4\sqrt{3} -2 + 3\sqrt{3} + 6$

and $\;$ $\left(\sqrt{3} - 2\right)x + \left(2 \sqrt{3} +1\right) y = 4\sqrt{3} +2 + 3\sqrt{3} - 6$

i.e. $\;$ $\left(\sqrt{3} + 2\right)x + \left(2 \sqrt{3} -1\right) y = 7\sqrt{3} + 4$ $\;\;\; \cdots \; (1)$

and $\;$ $\left(\sqrt{3} - 2\right)x + \left(2 \sqrt{3} +1\right) y = 7\sqrt{3} - 4$ $\;\;\; \cdots \; (2)$

Equations $(1)$ and $(2)$ are the required straight lines.

Equation $(1)$ cuts the $X$ axis at the point $A \left(x_1, y_1\right) = \left(\dfrac{7 \sqrt{3} + 4}{\sqrt{3} + 2}, 0\right)$

Equation $(2)$ cuts the $X$ axis at the point $B \left(x_2, y_2\right) = \left(\dfrac{7 \sqrt{3} - 4}{\sqrt{3} - 2}, 0\right)$

Comparing equation $(1)$ with the standard equation of a straight line $\;$ $a_1x + b_1y + c_1 = 0$ gives

$a_1 = \sqrt{3} + 2$, $\;$ $b_1 = 2 \sqrt{3} - 1$, $\;$ $c_1 = - 7 \sqrt{3} - 4$

Comparing equation $(2)$ with the standard equation of a straight line $\;$ $a_2x + b_2y + c_2 = 0$ gives

$a_2 = \sqrt{3} - 2$, $\;$ $b_2 = 2 \sqrt{3} + 1$, $\;$ $c_2 = - 7 \sqrt{3} + 4$

The $Y$ coordinate of the point of intersection of equations $(1)$ and $(2)$ is

$y_3 = \dfrac{c_1 a_2 - c_2 a_1}{a_1 b_2 - a_2 b_1}$

i.e. $\;$ $y_3 = \dfrac{- \left(7 \sqrt{3} + 4\right) \left(\sqrt{3} - 2\right) - \left(4 - 7 \sqrt{3}\right) \left(\sqrt{3} + 2\right)}{\left(\sqrt{3} + 2\right) \left(2 \sqrt{3} + 1\right) - \left(\sqrt{3} - 2 \right) \left(2 \sqrt{3} - 1\right)}$

i.e. $\;$ $y_3 = \dfrac{\left(-21 + 14 \sqrt{3} - 4 \sqrt{3} + 8\right) - \left(4 \sqrt{3} + 8 - 21 - 14 \sqrt{3}\right)}{\left(6 + \sqrt{3} + 4 \sqrt{3} + 2\right) - \left(6 - \sqrt{3} - 4 \sqrt{3} + 2\right)}$

i.e. $\;$ $y_3 = \dfrac{-13 + 10 \sqrt{3} + 10 \sqrt{3} + 13}{8 + 5 \sqrt{3} - 8 + 5 \sqrt{3}} = \dfrac{20 \sqrt{3}}{10 \sqrt{3}} = 2$

Length of base of triangle enclosed by the lines $(1)$ and $(2)$ and the $X$ axis is $= \left|AB\right|$

$AB = \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2}$

i.e. $\;$ $AB = \sqrt{\left(\dfrac{7 \sqrt{3} - 4}{\sqrt{3} - 2} - \dfrac{7 \sqrt{3} + 4}{\sqrt{3} + 2}\right)^2 + 0^2}$

i.e. $\;$ $AB = \dfrac{21 + 14 \sqrt{3} - 4 \sqrt{3} - 8 - 21 + 14 \sqrt{3} - 4 \sqrt{3} + 8}{\left(\sqrt{3}\right)^2 - 2^2}$

i.e. $\;$ $AB = \dfrac{20 \sqrt{3}}{-1}$

$\therefore \;$ Length of base of triangle $= \left|AB\right| = 20 \sqrt{3}$

Height of triangle $= y_3 = 2$

$\therefore \;$ Area of required triangle $= \dfrac{1}{2} \times \text{ base } \times \text{ height}$

i.e. $\;$ Area $= \dfrac{1}{2} \times 20 \sqrt{3} \times 2 = 20 \sqrt{3}$ sq units

Coordinate Geometry - Straight Line

Find the acute angle between the pair of lines $\;$ $2x + y + 4 = 0$ and $y - 3x = 7$

Let $\theta$ be the angle between the lines $\;$ $2x + y + 4 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and

$y - 3x = 7$ $\;\;\; \cdots \; (2)$

Equation $(1)$ can be written as $\;$ $y = -2x + 4$

$\therefore \;$ Slope of equation $(1)$ is $\;$ $m_1 = -2$

Equation $(2)$ can be written as $\;$ $y = 3x + 7$

$\therefore \;$ Slope of equation $(2)$ is $\;$ $m_2 = 3$

Now, $\;$ $\theta = \tan^{-1} \left(\dfrac{m_1 - m_2}{1 + m_1 m_2}\right)$

i.e. $\;$ $\theta = \tan^{-1} \left(\dfrac{-2 - 3}{1 + \left(-2\right) \times 3}\right)$

i.e. $\;$ $\theta = \tan^{-1} \left(\dfrac{-5}{-5}\right)$

i.e. $\;$ $\theta = \tan^{-1} \left(1\right) = 45^\circ$

Coordinate Geometry - Straight Line

$OACB$ is a rectangle where $OA = a$, $OB = b$. Taking $OA$, $OB$ as axes of coordinates, prove that the equation of $AE$ where $E$ is the midpoint of $BC$ is $\;$ $\dfrac{x}{a} + \dfrac{y}{2b} = 1$.

Given: $\;$ $OA$ and $OB$ are the axes of coordinates.

Let $OA$ represent the $X$ axis and $OB$ represent the $Y$ axis.

Let $O$ be the origin $\left(0,0\right)$.

Given: $\;$ $OA = a$ and $OB = b$ $\implies$ $A = \left(a, 0\right)$ and $B = \left(0, b\right)$

Then $\;$ $C = \left(a, b\right)$

$E$ is the midpoint of $BC$.

$\therefore \;$ The coordinates of $E = \left(\dfrac{a + 0}{2}, \dfrac{b + b}{2}\right) = \left(\dfrac{a}{2}, b\right)$

Slope of $AE = \dfrac{b - 0}{\dfrac{a}{2} - a} = \dfrac{-2b}{a}$

$\therefore \;$ Equation of $AE$ is

$y - 0 = \dfrac{-2b}{a} \left(x - a\right)$

i.e. $\;$ $ay = -2bx + 2ab$

i.e. $\;$ $\dfrac{y}{2b} = - \dfrac{x}{a} + 1$

i.e. $\;$ $\dfrac{x}{a} + \dfrac{y}{2b} = 1$

Coordinate Geometry - Straight Line

A straight line is drawn through the point $\left(\sqrt{3}, 2\right)$ and inclined at an angle of $30^\circ$ to the $X$ axis. Find the length intercepted on the line between the given point and the line $\sqrt{3}x + y = 9$.

The given line passes through the point $\left(\sqrt{3}, 2\right)$ and is inclined at an angle of $30^\circ$ to the $X$ axis.

$\therefore \;$ The equation of the line can be written as

$\dfrac{x - \sqrt{3}}{\cos 30^\circ} = \dfrac{y - 2}{\sin 30^\circ} = r$ $\;\;\; \cdots \; (1)$

where $\;$ $r$ $\;$ is the distance between $\left(\sqrt{3}, 2\right)$ and any point $P \left(x, y\right)$.

We have from equation $(1)$

$x = \sqrt{3} + r \cos 30^\circ = \sqrt{3} + \dfrac{\sqrt{3} r}{2}$

and $\;$ $y = 2 + r \sin 30^\circ = 2 + \dfrac{r}{2}$

If $P \left(x, y\right)$ be the point where the line given by equation $(1)$ meets the line $\;$ $\sqrt{3}x + y = 9$ $\;\;\; \cdots \; (2)$,

the coordinates of $P$ must satisfy equation $(2)$.

Substituting the values of $x$ and $y$ in equation $(2)$ gives

$\sqrt{3} \left(\sqrt{3} + \dfrac{\sqrt{3}r}{2}\right) + 2 + \dfrac{r}{2} = 9$

i.e. $\;$ $3 + \dfrac{3r}{2} + 2 + \dfrac{r}{2} = 9$

i.e. $\;$ $2r = 4$ $\implies$ $r = 2$

i.e. $\;$ the length intercepted $= 2$

Coordinate Geometry - Straight Line

Find the equation of a straight line joining the pair of points $\left(a \cos \theta, b \sin \theta\right)$ and $\left(a \cos \phi, b \sin \phi\right)$.

Let $\;$ $A \left(x_1, y_1\right) = \left(a \cos \theta, b \sin \theta\right)$, $\;$ $B \left(x_2, y_2\right) = \left(a \cos \phi, b \sin \phi\right)$

Slope of the required line $= m = \dfrac{y_2 - y_1}{x_2 - x_1}$

i.e. $\;$ $m = \dfrac{b \sin \phi - b \sin \theta}{a \cos \phi - a \cos \theta}$

i.e. $\;$ $m = \dfrac{b \left(\sin \phi - \sin \theta\right)}{a \left(\cos \phi - \cos \theta\right)}$

$\left[\text{Note: } \sin C - \sin D = 2 \cos \left(\dfrac{C + D}{2}\right) \sin \left(\dfrac{C - D}{2}\right) \right.$
$\left. \cos C - \cos D = - 2 \sin \left(\dfrac{C + D}{2}\right) \sin \left(\dfrac{C - D}{2}\right) \right]$

i.e. $\;$ $m = \dfrac{2b \cos \left(\dfrac{\theta + \phi}{2}\right) \sin \left(\dfrac{\phi - \theta}{2}\right)}{-2a \sin \left(\dfrac{\theta + \phi}{2}\right) \sin \left(\dfrac{\phi - \theta}{2}\right)}$

i.e. $\;$ $m = \dfrac{-b}{a} \times \dfrac{\cos \left(\dfrac{\theta + \phi}{2}\right)}{\sin \left(\dfrac{\theta + \phi}{2}\right)}$

Equation of the required line is of the form: $\;$ $y - y_1 = m \left(x - x_1\right)$

i.e. $\;$ $y - b \sin \theta = \dfrac{-b}{a} \times \dfrac{\cos \left(\dfrac{\theta + \phi}{2}\right)}{\sin \left(\dfrac{\theta + \phi}{2}\right)} \times \left(x - a \cos \theta\right)$

i.e. $\;$ $y \sin \left(\dfrac{\theta + \phi}{2}\right)a - ab \sin \theta \sin \left(\dfrac{\theta + \phi}{2}\right)$
$\hspace{1.5cm}$ $ = - b \cos \left(\dfrac{\theta + \phi}{2}\right)x + ab \cos \theta \cos \left(\dfrac{\theta + \phi}{2}\right)$

i.e. $\;$ $x \cos \left(\dfrac{\theta + \phi}{2}\right)b + y \sin \left(\dfrac{\theta + \phi}{2}\right)a$
$\hspace{1.5cm}$ $ = ab \left[\sin \theta \sin \left(\dfrac{\theta + \phi}{2}\right) + \cos \theta \cos \left(\dfrac{\theta + \phi}{2}\right)\right]$

$\left[\text{Note: } \cos \left(A - B\right) = \cos A \cos B + \sin A \sin B\right]$

i.e. $\;$ $\dfrac{x}{a} \cos \left(\dfrac{\theta + \phi}{2}\right) + \dfrac{y}{b} \sin \left(\dfrac{\theta + \phi}{2}\right) = \cos \left(\theta - \dfrac{\theta + \phi}{2}\right)$

i.e. $\;$ $\dfrac{x}{a} \cos \left(\dfrac{\theta + \phi}{2}\right) + \dfrac{y}{b} \sin \left(\dfrac{\theta + \phi}{2}\right) = \cos \left(\dfrac{\theta - \phi}{2}\right)$

Coordinate Geometry - Straight Line

Find the equation of a straight line joining the pair of points $\left(at_1^2, 2at_1\right)$ and $\left(at_2^2, 2at_2\right)$.

Let $\;$ $A \left(x_1, y_1\right) = \left(at_1^2, 2at_1\right)$, $\;$ $B \left(x_2, y_2\right) = \left(at_2^2, 2at_2\right)$

Slope of the required line $= m = \dfrac{y_2 - y_1}{x_2 - x_1}$

i.e. $\;$ $m = \dfrac{2at_2 - at_1}{at_2^2 -at_1^2}$

i.e. $\;$ $m = \dfrac{2a \left(t_2 - t_1\right)}{a \left(t_2^2 - t_1^2\right)}$

i.e. $\;$ $m = \dfrac{2 \left(t_2 - t_1\right)}{\left(t_2 - t_1\right) \left(t_2 + t_1\right)}$

i.e. $\;$ $m = \dfrac{2}{t_1 + t_2}$

Equation of the required line is of the form: $\;$ $y - y_1 = m \left(x - x_1\right)$

i.e. $\;$ $y - 2at_1 = \left(\dfrac{2}{t_1 + t_2}\right) \left(x - at_1^2\right)$

i.e. $\;$ $y \left(t_1 + t_2\right) - 2 a t_1 \left(t_1 + t_2\right) = 2x - 2a t_1^2$

i.e. $\;$ $2x - y \left(t_1 + t_2\right) + 2at_1 \left(t_1 + t_2\right) - 2 a t_1^2 = 0$

i.e. $\;$ $2x - y \left(t_1 + t_2\right) + 2 a t_1 \left(t_1 + t_2 - t_1\right) = 0$

i.e. $\;$ $2x - y \left(t_1 + t_2\right) + 2 a t_1 t_2 = 0$

Coordinate Geometry - Straight Line

Find the equation of a straight line which passes through the point $\left(-1, 2\right)$ and is inclined at $60^\circ$ to the $X$ axis.

Inclination of the required line to the $X$ axis $= \theta = 60^\circ$

$\therefore \;$ Slope of the required line $= m = \tan \theta = \tan 60^\circ = \sqrt{3}$

The required line passes through the point $\left(x_1, y_1\right) = \left(-1, 2\right)$

Equation of the required line is of the form: $\;$ $y - y_1 = m \left(x - x_1\right)$

$\therefore \;$ The equation of the required line is

$y - 2 = \sqrt{3} \left(x + 1\right)$

i.e. $\;$ $\sqrt{3} x - y + \sqrt{3} + 2 = 0$

Coordinate Geometry - Straight Line

Reduce the equation $\;$ $3x + 4y = 5$ $\;$ to the slope-intercept form, the intercept form and the perpendicular form.

Equation of given straight line: $\;$ $3x + 4y = 5$ $\;\;\; \cdots \; (1)$

Slope-intercept form:

Equation $(1)$ can be written as

$4y = -3x + 5$

i.e. $\;$ $y = \dfrac{-3}{4} x + \dfrac{5}{4}$ $\;\;\; \cdots \; (2)$

Equation $(2)$ is the slope-intercept form of equation $(1)$ with slope $= m = \dfrac{-3}{4}$ and intercept $= c = \dfrac{5}{4}$

Intercept form:

Rearranging equation $(1)$ we have,

$\dfrac{3x}{5} + \dfrac{4 y}{5} = 1$

i.e. $\;$ $\dfrac{x}{5 / 3} + \dfrac{y}{5 / 4} = 1$ $\;\;\; \cdots \; (3)$

Equation $(3)$ is the intercept form of equation $(1)$ with intercept on the $X$ axis $= a = \dfrac{5}{3}$ and intercept on the $Y$ axis $= b = \dfrac{5}{4}$

Perpendicular form:

Writing equation $(1)$ as

$3x + 4y - 5 = 0$

i.e. $\;$ $\dfrac{3x}{\sqrt{3^2 + 4^2}} + \dfrac{4y}{\sqrt{3^2 + 4^2}} - \dfrac{5}{\sqrt{3^2 + 4^2}} = 0$

i.e. $\;$ $\dfrac{3x}{5} + \dfrac{4y}{5} - \dfrac{5}{5} = 0$

i.e. $\;$ $x \cos \alpha + y \sin \alpha -1 = 0$ $\;\;\; \cdots \; (4)$

where $\;$ $\cos \alpha = \dfrac{3}{5}$, $\;$ $\sin \alpha = \dfrac{4}{5}$ $\;$ i.e. $\;$ $\tan \alpha = \dfrac{4}{3}$

Equation $(4)$ is the required perpendicular form of equation $(1)$.

Coordinate Geometry - Straight Line

Find the equation of the straight line which passes through the point $\left(-5, 4\right)$ and is such that the portion of it intercepted between the axes is divided by the point in the ratio $1 : 2$

Let

intercept made by the required line on the $X$ axis $= a$

intercept made by the required line on the $Y$ axis $= b$

Then, the required line cuts the $X$ axis at the point $A \left(a, 0\right)$ and the $Y$ axis at the point $B \left(0, b\right)$.

The required equation passes through the point $P\left(-5, 4\right)$.

Given: $\;$ The point $P$ divides the line join of points $A$ and $B$ internally in the ratio $1 : 2$.

$\therefore \;$ By section formula for internal division, we have,

for the $x$ coordinate: $\;\;$ $-5 = \dfrac{1 \times 0 + 2 \times a}{1 + 2}$ $\implies$ $- 15 = 2a$ $\implies$ $a = \dfrac{-15}{2}$

and for the $y$ coordinate: $\;\;$ $4 = \dfrac{1 \times b + 2 \times 0}{1 + 2}$ $\implies$ $b = 12$

The equation of the required line be of the form: $\;$ $\dfrac{x}{a} + \dfrac{y}{b} = 1$

$\dfrac{x}{-15 / 2} + \dfrac{y}{12} = 1$

i.e. $\;$ $\dfrac{2x}{-15} + \dfrac{y}{12} = 1$

i.e. $\;$ $24x - 15y = - 180$

$\therefore \;$ The equation of the required line is

$8x - 5y + 60 = 0$

Coordinate Geometry - Straight Line

Find the equation of the straight line which passes through the point $\left(2, -3\right)$ and cuts off intercepts on the axes equal in magnitude and of the same sign.

Let

intercept made by the required line on the $X$ axis $= a$

intercept made by the required line on the $Y$ axis $= b$

the equation of the required line be of the form: $\;$ $\dfrac{x}{a} + \dfrac{y}{b} = 1$

Given: $\;$ $a = b$

$\therefore \;$ The equation of the required line becomes $\;$ $\dfrac{x}{a} + \dfrac{y}{a} = 1$

i.e. $\;$ $x + y = a$ $\;\;\; \cdots \; (1)$

The required equation passes through the point $\left(2, -3\right)$.

Substituting the point in equation $(1)$ gives

$2 - 3 = a$ $\implies$ $a = -1$

Substituting the value of $a$ in equation $(1)$ gives

$x + y = -1$

$\therefore \;$ The equation of the required line is $\;$ $x + y + 1 = 0$

Coordinate Geometry - Straight Line

Find the equation of the straight line which cuts off an intercept $3$ from the $Y$ axis and is inclined at an angle $\tan^{-1} \left(\dfrac{2}{3}\right)$ to the $X$ axis.

Intercept on the $Y$ axis $= c = 3$

Inclination of the required line to the $X$ axis $= \theta = \tan^{-1} \left(\dfrac{2}{3}\right)$

$\therefore \;$ Slope of the required line $= m = \tan \theta = \tan \left[\tan^{-1} \left(\dfrac{2}{3}\right)\right] = \dfrac{2}{3}$

The equation of the required line is of the form: $\;$ $y = mx + c$

i.e. $\;$ $y = \dfrac{2}{3} \times x + 3$

$\therefore \;$ The equation of the required line is: $\;$ $2x -3y + 9 = 0$

Coordinate Geometry - Straight Line

Find the equation of the straight line which cuts off an intercept $2$ from the positive direction of the $Y$ axis and is inclined at $45^\circ$ to the positive direction of the $X$ axis.

Intercept from the positive direction of the $Y$ axis $= c = 2$

Inclination of the required line to the positive direction of the $X$ axis $= \theta = 45^\circ$

$\therefore \;$ Slope of the required line $= m = \tan \theta = 1$

The equation of the required line is of the form: $\;$ $y = mx + c$

i.e. $\;$ $y = 1 \times x + 2$

$\therefore \;$ The equation of the required line is: $\;$ $y = x + 2$

Coordinate Geometry - Locus

Find the equation of the locus of a point such that the sum of the squares of whose distances from the points $\left(2, 4\right)$ and $\left(-3, -1\right)$ is $30$.

Let $\;$ $A = \left(2, 4\right)$, $\;$ $B = \left(-3, -1\right)$

Let $P \left(x, y\right)$ be a point on the required locus.

Distance $\;$ $PA = \sqrt{\left(x - 2\right)^2 + \left(y - 4\right)^2}$

i.e. $\;$ $PA = \sqrt{x^2 + y^2 - 4x - 8y + 20}$

$\therefore \;$ $\left(PA\right)^2 = x^2 + y^2 - 4x - 8y + 20$

Distance $\;$ $PB = \sqrt{\left(x + 3\right)^2 + \left(y + 1\right)^2}$

i.e. $\;$ $PB = \sqrt{x^2 + y^2 + 6x + 2y + 10}$

$\therefore \;$ $\left(PB\right)^2 = x^2 + y^2 + 6x + 2y + 10$

As per question, $\;$ $\left(PA\right)^2 + \left(PB\right)^2 = 30$

i.e. $\;$ $x^2 + y^2 - 4x - 8y + 20 + x^2 + y^2 + 6x + 2y + 10 = 30$

i.e. $\;$ $2 x^2 + 2 y^2 + 2x - 6y = 0$

i.e. $\;$ $x^2 + y^2 + x - 3y = 0$ $\;\;\;$ is the equation of the required locus.

Coordinate Geometry - Locus

Find the equation of the locus of a point whose distance from $\left(a, 0\right)$ is always four times its distance from the $Y$ axis.

Let $\;$ $A = \left(a, 0\right)$

Let $P \left(x, y\right)$ be a point on the required locus.

Distance $\;$ $PA = \sqrt{\left(x - a\right)^2 + \left(y - 0\right)^2}$

i.e. $\;$ $PA = \sqrt{x^2 + y^2 + a^2 - 2ax}$

$\therefore \;$ $\left(PA\right)^2 = x^2 + y^2 + a^2 - 2ax$

Distance of point $P \left(x, y\right)$ from the $Y$ axis is $= x$

As per question, $\;$ $PA = 4 \times x$

i.e. $\;$ $\left(PA\right)^2 = 16 \times x^2$

i.e. $\;$ $x^2 + y^2 + a^2 - 2ax = 16x^2$

i.e. $\;$ $15x^2 - y^2 + 2ax - a^2 = 0$ $\;\;\;$ is the equation of the required locus.

Coordinate Geometry - Locus

Find the equation of the locus of a point whose distance from $\left(2, -3\right)$ is double its distance from $\left(1, 2\right)$.

Let the given points be $\;$ $A \left(2, -3\right)$ $\;$ and $\;$ $B \left(1, 2\right)$.

Let $P \left(x, y\right)$ be a point on the given locus.

Distance $\;$ $PA = \sqrt{\left(x - 2\right)^2 + \left(y + 3\right)^2}$

Distance $\;$ $PB = \sqrt{\left(x - 1\right)^2 + \left(y - 2\right)^2}$

As per question, $\;$ $PA = 2 \times PB$

i.e. $\;$ $\left(PA\right)^2 = 4 \times \left(PB\right)^2$

i.e. $\;$ $\left(x - 2\right)^2 + \left(y + 3\right)^2 = 4 \times \left[\left(x - 1\right)^2 + \left(y - 2\right)^2\right]$

i.e. $\;$ $x^2 - 4x + 4 + y^2 + 6y + 9 = 4 \times \left[x^2 - 2x + 1 + y^2 - 4y + 4\right]$

i.e. $\;$ $x^2 + y^2 -4x + 6y + 13 = 4x^2 + 4y^2 -8x - 16y + 20$

i.e. $\;$ $3x^2 + 3y^2 -4x -22y +7 = 0$ $\;\;\;$ is the equation of the required locus.

Coordinate Geometry - Locus

Find the equation of the locus of points which are equidistant from the points $\left(2, 3\right)$ and $\left(5, 7\right)$.

Let the given points be $\;$ $A \left(2, 3\right)$ $\;$ and $\;$ $B \left(5, 7\right)$.

Let $P \left(x, y\right)$ be a point on the given locus.

As per question, $\;$ $PA = PB$

i.e. $\;$ $\left(PA\right)^2 = \left(PB\right)^2$

i.e. $\;$ $\left(x - 2\right)^2 + \left(y - 3\right)^2 = \left(x - 5\right)^2 + \left(y - 7\right)^2$

i.e. $\;$ $x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 - 10x + 25 + y^2 - 14y + 49$

i.e. $\;$ $6x + 8y - 61 = 0$ $\;\;\;$ is the equation of the required locus.

Coordinate Geometry - Cartesian Coordinates

If the point $D$ divides the base $BC$ of a $\triangle ABC$ in the ratio $n : m$, prove that

$m \cdot \left(AB\right)^2 + n \cdot \left(AC\right)^2 = \left(m + n\right) \left(AD\right)^2 + m \cdot \left(BD\right)^2 + n \cdot \left(DC\right)^2$

Let $B$ be at the origin. Then the coordinates of point $B = \left(0,0\right)$

Let $BC$ be along the $X$ axis and the line perpendicular to $BC$ be the $Y$ axis.

Let $BC = a$. $\;$ Then $\;$ $C = \left(a, 0\right)$

Let $\;$ $a = \left(x, y\right)$

Since point $D$ divides the base $BC$, point $D$ is also on the $X$ axis.

Let $\;$ $D = \left(p, 0\right)$

The point $D$ divides $BC$ in the ratio $n : m$

$\therefore \;$ By section formula for internal division, the $x$ coordinate of point $D$ is

$p = \dfrac{n \times a + m \times 0}{m + n} = \dfrac{na}{m + n}$

$\therefore \;$ $D = \left(\dfrac{na}{m + n}, 0\right)$

Now, by distance formula,

$\left(AB\right)^2 = \left(x - 0\right)^2 + \left(y - 0\right)^ = x^2 + y^2$

$\left(AC\right)^2 = \left(x - a\right)^2 + \left(y - 0\right)^2 = x^2 + a^2 - 2ax + y^2$

$\left(AD\right)^2 = \left(x - \dfrac{na}{m + n}\right)^2 + \left(y - 0\right)^2 = x^2 + \dfrac{a^2 n^2}{\left(m + n\right)^2} - \dfrac{2xna}{m + n} + y^2$

$\left(BD\right)^2 = \left(0 - \dfrac{na}{m + n}\right)^2 + 0^2 = \dfrac{a^2 n^2}{\left(m + n\right)^2}$

$\left(DC\right)^2 = \left(a - \dfrac{na}{m + n}\right)^2 + 0^2 = \left(\dfrac{am + an - an}{m + n}\right)^2 = \dfrac{a^2 m^2}{\left(m + n\right)^2}$

Now,

$\begin{aligned} LHS & = m \cdot \left(AB\right)^2 + n \cdot \left(AC\right)^2 \\\\ & = m \left(x^2 + y^2\right) = n \left(x^2 + a^2 - 2ax + y^2\right) \\\\ & = x^2 \left(m + n\right) + y^2 \left(m + n\right) + a^2 n - 2anx \\\\ & = \left(m + n\right) \left(x^2 + y^2\right) + an \left(a - 2x\right) \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} RHS & = \left(m + n\right) \left(AD\right)^2 + m \cdot \left(BD\right)^2 + n \cdot \left(DC\right)^2 \\\\ & = \left(m + n\right) \left[x^2 + \dfrac{a^2 n^2}{\left(m + n\right)^2} - \dfrac{2xna}{m + n} + y^2\right] + m \times \dfrac{a^2 n^2}{\left(m + n\right)^2} + n \times \dfrac{a^2 m^2}{\left(m + n\right)^2} \\\\ & = \left(m + n\right) x^2 + \left(m + n\right) y^2 + \dfrac{a^2 n^2}{m + n} - 2xna + \dfrac{a^2}{\left(m + n\right)^2} \left(mn^2 + nm^2\right) \\\\ & = \left(m + n\right) \left(x^2 + y^2\right) + \dfrac{a^2 n^2}{m + n} - 2xna + \dfrac{a^2 mn}{\left(m + n\right)^2} \left(n + m\right) \\\\ & = \left(m + n\right) \left(x^2 + y^2\right) - 2xna + \dfrac{a^2 n^2}{m + n} + \dfrac{a^2 mn}{m + n} \\\\ & = \left(m + n\right) \left(x^2 + y^2\right) - 2xna + a^2 n \left(\dfrac{n}{n + m} + \dfrac{m}{m + n}\right) \\\\ & = \left(m + n\right) \left(x^2 + y^2\right) - 2xna + a^2 n \\\\ & = \left(m + n\right) \left(x^2 + y^2\right) + an \left(a - 2x\right) \;\;\; \cdots \; (2) \end{aligned}$

$\therefore \;$ From equations $(1)$ and $(2)$ we have $LHS = RHS$.

Hence proved.

Coordinate Geometry - Cartesian Coordinates

The centroid of triangle $ABC$ is at the point $\left(2, 3\right)$. The coordinates of $A$ and $B$ are $\left(5, 6\right)$ and $\left(-1, 4\right)$. Find the coordinates of point $C$.

Given: $\;$ $A \left(x_1, y_1\right) = \left(5, 6\right)$, $\;$ $B \left(x_2, y_2\right) = \left(-1, 4\right)$, $\;$ centroid $G \left(p, q\right) = \left(2, 3\right)$

Let the coordinates of point $C = \left(x, y\right)$

Let $M$ be the midpoint of $AB$.

Then, by midpoint formula, the coordinates of point $M$ are

$M = \left(\dfrac{5 - 1}{2}, \dfrac{6 + 4}{2}\right) = \left(2, 5\right)$

$CM$ is the median of $\triangle ABC$.

The centroid $G$ divides the median $CM$ internally in the ratio $2 : 1$.

$\therefore \;$ By section formula for internal division, we have,

for the $x$ coordinate: $\;$ $2 = \dfrac{2 \times 2 + 1 \times x}{2 + 1}$

i.e. $\;$ $6 = 4 + x$ $\implies$ $x = 2$

for the $y$ coordinate: $\;$ $3 = \dfrac{2 \times 5 + 1 \times y}{2 + 1}$

i.e. $\;$ $9 = 10 + y$ $\implies$ $y = -1$

$\therefore \;$ $C \left(x, y\right) = \left(2, -1\right)$

Coordinate Geometry - Cartesian Coordinates

Find the area of the quadrilateral whose angular points are $\left(-1, 6\right)$, $\left(-3, -9\right)$, $\left(5, -8\right)$ and $\left(3, 9\right)$.

Let $\;$ $A \left(x_1, y_1\right) = \left(-1, 6\right)$, $\;$ $B \left(x_2, y_2\right) = \left(-3, -9\right)$, $\;$ $C \left(x_3, y_3\right) = \left(5, -8\right)$, $\;$ $D \left(x_4, y_4\right) = \left(3, 9\right)$

Area of quadrilateral $ABCD$ is

$\begin{aligned} \Delta & = \dfrac{1}{2} \left(x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1 - y_1 x_2 - y_2 x_3 - y_3 x_4 - y_4 x_1\right) \\\\ & = \dfrac{1}{2} \left[-1 \times \left(-9\right) - 3 \times \left(-8\right) + 5 \times 9 + 3 \times 6 -6 \times \left(-3\right) + 9 \times 5 + 8 \times 3 - 9 \times \left(-1\right) \right] \\\\ & = \dfrac{1}{2} \left[9 + 24 + 45 + 18 + 18 + 45 + 24 + 9\right] \\\\ & = \dfrac{1}{2} \times 192 = 96 \end{aligned}$

Coordinate Geometry - Cartesian Coordinates

Show that the points $\left(a, b + c\right)$, $\left(b, c + a\right)$ and $\left(c, a + b\right)$ are collinear.

Let $A \left(x_1, y_1\right) = \left(a, b + c\right)$, $\;$ $B \left(x_2, y_2\right) = \left(b, c + a\right)$ and $C \left(x_3, y_3\right) = \left(c, a + b\right)$

Area of triangle formed by the points $ABC$ is

$\begin{aligned} \Delta & = \dfrac{1}{2} \left[x_1 \left(y_2 - y_3\right) + x_2 \left(y_3 - y_1\right) + x_3 \left(y_1 - y_2\right)\right] \\\\ & = \dfrac{1}{2} \left[a \left(c + a - a - b\right) + b \left(a + b - b - c\right) + c \left(b + c -c - a\right)\right] \\\\ & = \dfrac{1}{2} \left[a \left(c - b\right) + b \left(a - c\right) + c \left(b - a\right)\right] \\\\ & = \dfrac{1}{2} \left[ac - ab + ab - bc + bc - ac\right] \\\\ & = 0 \end{aligned}$

$\because \;$ The area of triangle formed by the points $A$, $B$ and $C$ is zero,

$\implies$ the points $A$, $B$ and $C$ are collinear.

Coordinate Geometry - Cartesian Coordinates

Find the area of the triangle whose vertices are $\left(a \cos \theta , b \sin \theta\right)$, $\left(-a \sin \theta, b \cos \theta\right)$ and $\left(-a \cos \theta, -b \sin \theta\right)$

Let $\;$ $A \left(x_1, y_1\right) = \left(a \cos \theta , b \sin \theta\right)$, $\;$ $B \left(x_2, y_2\right) = \left(-a \sin \theta, b \cos \theta\right)$, $\;$ $C \left(x_3, y_3\right) = \left(-a \cos \theta, -b \sin \theta\right)$

Area of $\triangle ABC = \dfrac{1}{2} \left[x_1 \left(y_2 - y_3\right) + x_2 \left(y_3 - y_1\right) + x_3 \left(y_1 - y_2\right)\right]$

$= \dfrac{1}{2} \left[a \cos \theta \left(b \cos \theta + b \sin \theta\right) - a \sin \theta \left(-b \sin \theta - b \sin \theta\right) - a \cos \theta \left(b \sin \theta - b \cos \theta\right)\right]$

$= \dfrac{1}{2} \left[ab \cos^2 \theta + ab \sin \theta \cos \theta + 2ab \sin^2 \theta -ab \sin \theta \cos \theta + ab \cos^2 \theta\right]$

$= \dfrac{1}{2} \left[2ab \cos^2 \theta + 2 ab \sin^2 \theta\right]$

$= \dfrac{1}{2} \left[2ab \left(\sin^2 \theta + \cos^2 \theta\right)\right]$

$= \dfrac{1}{2} \times 2ab \times 1 = ab$

Coordinate Geometry - Cartesian Coordinates

$P \left(3,7\right)$ is a point on the line joining $A \left(1,1\right)$ and $B \left(6,16\right)$. Find the harmonic conjugate of $P$ with respect to $A$ and $B$.

Let the point $P \left(3,7\right)$ divide the line joining $A \left(1,1\right)$ and $B \left(6,16\right)$ internally in the ratio $k : 1$.

Then, by section formula,

$3 = \dfrac{1 \times 1 + 6 \times k}{k + 1}$

i.e. $\;$ $3k + 3 = 1 + 6k$

i.e. $\;$ $3k = 2$ $\implies$ $k = \dfrac{2}{3}$

i.e. $\;$ the point $P$ divides the line $AB$ internally in the ratio $2:3$

Let $Q \left(x,y\right)$ be the harmonic conjugate of $P$ with respect to $A$ and $B$.

Then the point $Q$ divides $AB$ externally in the ratio $2 : 3$

Then, by section formula for external division,

$x = \dfrac{2 \times 6 - 3 \times 1}{2 - 3} = \dfrac{12 - 3}{-1} = -9$

$y = \dfrac{2 \times 16 - 3 \times 1}{2 - 3} = \dfrac{32 - 3}{-1} = -29$

$\therefore \;$ $Q \left(x, y\right) = \left(-9, -29\right)$

Coordinate Geometry - Cartesian Coordinates

The point $R \left(22,23\right)$ divides the line join of $P \left(7,5\right)$ and $Q$ externally in the ratio $3 : 5$. Find $Q$.

Let $Q = \left(p,q\right)$

The point $R \left(22,23\right)$ divides the line joining $P \left(7,5\right)$ and $Q \left(p,q\right)$ externally in the ratio $3:5$.

Then, by section formula,

$22 = \dfrac{p \times 3 - 7 \times 5}{3 - 5}$

i.e. $\;$ $22 = \dfrac{3p - 35}{-2}$

i.e. $\;$ $3p = -44 + 35$ $\implies$ $p = \dfrac{-9}{3} = -3$

and $\;$ $23 = \dfrac{q \times 3 - 5 \times 5}{3 - 5}$

i.e. $\;$ $23 = \dfrac{3q - 25}{-2}$

i.e. $\;$ $3q = - 46 + 25$ $\implies$ $q = \dfrac{-21}{3} = -7$

$\therefore \;$ $Q = \left(-3, -7\right)$

Coordinate Geometry - Cartesian Coordinates

Find the coordinates of the points which divide internally and externally the line joining $\left(-4, 4\right)$ and $(1,7)$ in the ratio $2 : 1$.

Let $A = \left(-4,4\right)$ and $B = \left(1,7\right)$

Let $P \left(p,q\right)$ divide the line joining $AB$ internally in the ratio $2:1$.

Then, by section formula,

$p = \dfrac{1 \times 2 + \left(-4\right) \times 1}{2 + 1} = \dfrac{2 - 4}{3} = \dfrac{-2}{3}$

$q = \dfrac{7 \times 2 + 4 \times 1}{2 + 1} = \dfrac{14 + 4}{3} = \dfrac{18}{3} = 6$

$\therefore \;$ $P = \left(\dfrac{-2}{3}, 6\right)$

Let $Q \left(x,y\right)$ divide the line joining $AB$ externally in the ratio $2:1$.

Then, by section formula,

$x = \dfrac{1 \times 2 - \left(-4\right) \times 1}{2 - 1} = \dfrac{2 + 4}{1} = 6$

$y = \dfrac{7 \times 2 - 4 \times 1}{2 - 1} = \dfrac{14 - 4}{1} = 10$

$\therefore \;$ $Q = \left(6, 10\right)$

Statistics - Ogive

Draw an ogive for the given frequency distribution:

Class Interval	$5 - 10$	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$	$30 - 35$
Frequency	$3$	$4$	$6$	$9$	$7$	$1$

Hence find the lower quartile, upper quartile and the inter-quartile range.

Class Interval	Frequency	Cumulative Frequency
$5 - 10$	$3$	$3$
$10 - 15$	$4$	$7$
$15 - 20$	$6$	$13$
$20 - 25$	$9$	$22$
$25 - 30$	$7$	$29$
$30 - 35$	$1$	$30$

Total number of observations $= N = 30$

Lower quartile $= Q_1 = \left(\dfrac{N}{4}\right)^{th}$ term $= \left(\dfrac{30}{4}\right)^{th}$ term $= 7.5^{th}$ term $= 15.5$

Upper quartile $= Q_3 = \left(\dfrac{3N}{4}\right)^{th}$ term $= \left(\dfrac{3 \times 30}{4}\right)^{th}$ term $= 22.5^{th}$ term $= 25.5$

Inter-quartile range $= Q_3 - Q_1 = 25.5 - 15.5 = 10$

Commercial Mathematics - Shares and Dividends

A person invests ₹ $9000$ in shares of a company which is paying $8 \%$ dividend. If ₹ $100$ shares are available at a discount of $10 \%$, find the number of shares purchased and the annual income.

Money invested $= $ ₹ $9000$

Face value (FV) of each share $= $ ₹ $100$

Market value (MV) of each share $= $ ₹ $100 - 10 \%$ of ₹ $100$

$ = $ ₹ $\left(100 - \dfrac{10}{100} \times 100\right) = $ ₹ $90$

$\therefore \;$ Number of shares bought $= \dfrac{\text{money invested}}{\text{MV of each share}} = \dfrac{9000}{90} = 100$

Dividend (income) on one share $= 8 \%$ of FV $= \dfrac{8}{100} \times $ ₹ $100 = $ ₹ $8$

$\therefore \;$ Total income from the shares $= 100 \times $ ₹ $8 = $ ₹ $800$

Probability

In a single throw of two dice, what is the probability of getting

1. a total of $9$;
2. a doublet;
3. $5$ on one die and $6$ on the other.

Let $S =$ event of throwing two dice.

Number of elements in sample space $= n \left(S\right) = 36$

1. Let $A =$ event of getting a total of $9$ when two dice are thrown
   
   i.e. $\;$ $A = \left\{\left(3,6\right), \left(4,5\right), \left(5,4\right), \left(6,3\right) \right\}$
   
   $\therefore \;$ Number of elements in $A = n \left(A\right) = 4$
   
   $\therefore \;$ Probability of event $A = P \left(A\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{4}{36} = \dfrac{1}{9}$



2. Let $B =$ event of getting a doublet when two dice are thrown
   
   i.e. $\;$ $B = \left\{\left(1,1\right), \left(2,2\right), \left(3,3\right), \left(4,4\right), \left(5,5\right), \left(6,6\right) \right\}$
   
   $\therefore \;$ Number of elements in $B = n \left(B\right) = 6$
   
   $\therefore \;$ Probability of event $B = P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{6}{36} = \dfrac{1}{6}$



3. Let $C =$ event of getting a $5$ on one die and $6$ on the other
   
   i.e. $\;$ $C = \left\{\left(5,6\right), \left(6,5\right) \right\}$
   
   $\therefore \;$ Number of elements in $C = n \left(C\right) = 2$
   
   $\therefore \;$ Probability of event $C = P \left(C\right) = \dfrac{n \left(C\right)}{n \left(S\right)} = \dfrac{2}{36} = \dfrac{1}{18}$

Circle

In the figure, $\angle BAD = 60^\circ$, $\;$ $\angle ABD = 70^\circ$, $\;$ $\angle BDC = 40^\circ$

1. Prove that $AC$ is the diameter of the circle.
2. Find $\angle ACB$.
3. Find $\angle BCD$.

$ABCD$ is a cyclic quadrilateral.

$\therefore \;$ $\angle BAD + \angle BCD = 180^\circ$ $\;\;\;$ [opposite angles of a cyclic quadrilateral are supplementary]

$\therefore \;$ $\angle BCD = 180^\circ - \angle BAD = 180^\circ - 60^\circ = 120^\circ$

In $\triangle BAD$, $\;$ $\angle BAD + \angle ABD + \angle ADB = 180^\circ$ $\;\;\;$ [sum of angles of a triangle]

$\therefore \;$ $\angle ADB = 180^\circ - \left(\angle BAD + \angle ABD\right) = 180^\circ - \left(60^\circ + 70^\circ\right) = 50^\circ$

Now, $\angle ADC = \angle ADB + \angle BDC = 50^\circ + 40^\circ = 90^\circ$

But angle in a semicircle is a right angle.

$\therefore \;$ $\angle ADC$ is an angle in a semicircle.

$\implies$ $AC$ is the diameter of the circle.

Now, $\;$ $\angle ADB = \angle ACB$ $\;\;\;$ [angles of same segment are equal]

$\therefore \;$ $\angle ACB = 50^\circ$

Similarity

In the figure, $AB$ and $DE$ are perpendicular to $BC$.

1. Prove that $\triangle ABC \sim \triangle DEC$.
2. If $AB = 6 \; cm$, $DE = 4 \; cm$, $AC = 15 \; cm$, calculate $CD$.
3. Find the ratio of $\;\;$ $\dfrac{\text{Area}\left(\triangle ABC\right)}{\text{Area}\left(\triangle DEC\right)}$

1. Given: $\;\;$ $AB \perp BC$, $\;$ $DE \perp BC$
   
   $\therefore \;$ $\angle ABC = \angle DEC = 90^\circ$ $\;\;\; \cdots \; (1)$
   
   In triangles $ABC$ and $DEC$,
   
   $\angle ACB = \angle DCE$ $\;$ [common angle] $\;\;\; \cdots \; (2)$
   
   $\therefore \;$ From equations $(1)$ and $(2)$, $\;$ $\triangle ABC \sim \triangle DEC$ $\;$ [by Angle-Angle postulate]



2. $\because \;$ $\triangle ABC \sim \triangle DEC$
   
   $\therefore \;$ $\dfrac{AB}{DE} = \dfrac{AC}{DC}$ $\;\;\;$ [corresponding sides of similar triangles are in proportion]
   
   $\therefore \;$ $DC = \dfrac{AC \times DE}{AB} = \dfrac{15 \times 4}{6} = 10 \; cm$



3. $\dfrac{\text{Area} \left(\triangle ABC\right)}{\text{Area}\left(\triangle DEC\right)} = \dfrac{AB^2}{DE^2}$
   
   [areas of two similar triangles are proportional to the squares on their corresponding sides]
   
   i.e. $\;$ $\dfrac{\text{Area} \left(\triangle ABC\right)}{\text{Area}\left(\triangle DEC\right)} = \dfrac{6^2}{4^2} = \dfrac{9}{4}$

Circle

In the figure, $O$ is the center of a circle.

$\angle ABC = 100^\circ$, $\angle ACD = 40^\circ$

$CT$ is a tangent to the circle at $C$.

Find $\angle ADC$ and $\angle DCT$.

$O$ is the center of the circle; $\;$ $\angle ABC = 100^\circ$; $\;$ $\angle ACD = 40^\circ$;

$CT$ is a tangent to the circle at a point $C$

In the figure, $\;$ $ABCD$ is a cyclic quadrilateral.

$\therefore \;$ $\angle ABC + \angle ADC = 180^\circ$ $\;\;\;$ [opposite angles of a cyclic quadrilateral are supplementary]

$\implies$ $\angle ADC = 180^\circ - \angle ABC = 180^\circ - 100^\circ = 80^\circ$

Now, $\;$ $\angle AOC = 2 \angle ADC$

[angle at the center of a circle is twice the angle at the remaining circumference]

$\therefore \;$ $\angle AOC = 2 \times 80^\circ = 160^\circ$

Join $OA$ and $OC$.

Then, $\;$ $OA = OC$ $\;\;\;$ [radii of the same circle]

$\implies$ $\angle ACO = \angle CAO$ $\;\;\;$ [angles opposite equal sides are equal]

In $\triangle AOC$, $\;$ $\angle AOC + \angle ACO + \angle CAO = 180^\circ$ $\;\;\;$ [sum of angles of a triangle]

i.e. $\;$ $\angle AOC + 2 \angle ACO = 180^\circ$

$\implies$ $\angle ACO = \dfrac{180^\circ - \angle AOC}{2} = \dfrac{180^\circ - 160^\circ}{2} = 10^\circ$

From the figure, $\;$ $\angle ACD = \angle ACO + \angle OCD$

$\therefore \;$ $\angle OCD = \angle ACD - \angle ACO = 40^\circ - 10^\circ = 30^\circ$

Also, $\;$ $\angle OCT = 90^\circ$

[tangent at any point of a circle and the radius through this point are perpendicular to each other]

But, $\;$ $\angle OCT = \angle OCD + \angle DCT$

$\therefore \;$ $\angle DCT = \angle OCT - \angle OCD = 90^\circ - 30^\circ = 60^\circ$

Mensuration

A hemispherical and a conical hole are scooped out of a solid wooden cylinder. The height of the solid cylinder is $7 \; cm$, radius of each of hemisphere, cone and cylinder is $3 \; cm$. The height of the cone is $3 \; cm$. Find the volume of the remaining solid. Give your answer correct to the nearest whole number. Take $\pi = 3.142$

Height of cylinder $= H = 7 \; cm$

Radius of cylinder $= R = 3 \; cm$

Radius of hemisphere $= R_S = 3 \; cm$

Radius of cone $= r = 3 \; cm$

Height of cone $= h = 3 \; cm$

Volume of cylinder $= V = \pi R^2 H$

Volume of hemisphere $= V_S = \dfrac{2}{3} \pi R_S^3$

Volume of cone $= V_C = \dfrac{1}{3} \pi r^2 h$

Volume of remaining solid $= V - \left(V_S + V_C\right)$

$= \pi R^2 H - \left(\dfrac{2}{3} \pi R_S^3 + \dfrac{1}{3} \pi r^2 h\right)$

$= \pi \left[3^2 \times 7 - \left(\dfrac{2}{3} \times 3^3 + \dfrac{1}{3} \times 3^2 \times 3\right)\right]$

$= \pi \left[63 - \left(18 + 9\right)\right] = 3.142 \times 36 = 113.112$

$\therefore \;$ Volume of the remaining solid $= 113 \; cm^3$ $\;\;\;$ [correct to the nearest whole number]

Constructions

Use a ruler and compass only for this question.

Draw a circle of radius $4 \; cm$.

Mark the center as $O$. Mark a point $P$ outside the circle at a distance of $7 \; cm$ from the center.

Construct two tangents to the circle from the external point $P$.

Measure and write down the length of the tangents.

Draw a circle with center $O$ and radius $4 \; cm$

Mark point $P$ at a distance of $7 \; cm$ from the center $O$.

Join $PO$.

Draw the perpendicular bisector of $OP$.

Draw a circle with $OP$ as diameter which cuts the given circle at points $A$ and $B$.

Join $PA$ and $PB$.

$PA$ and $PB$ are the required tangents.

Length of tangent $PA = PB = 5.7 \; cm$

Quadratic Equations

Solve the quadratic equation $\;$ $2x^2 - 5x - 10 = 0$ $\;$ giving your answer correct to two decimal places.

Given quadratic equation is $\;$ $2x^2 - 5x - 10 = 0$

Comparing with the standard quadratic equation $\;$ $ax^2 + bx + c = 0$ $\;$ gives

$a = 2$, $\;$ $b = -5$, $\;$ $c = -10$

Solution of the standard quadratic equation is $\;$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\therefore \;$ Solution of the given quadratic equation is

$\begin{aligned} x & = \dfrac{5 \pm \sqrt{5^2 - 4 \times 2 \times \left(-10\right)}}{2 \times 2} \\\\ & = \dfrac{5 \pm \sqrt{25 + 80}}{4} \\\\ & = \dfrac{5 \pm \sqrt{105}}{4} \\\\ & = \dfrac{5 \pm 10.247}{4} \\\\ & = \dfrac{15.247}{4} \;\;\; or \;\;\; \dfrac{-5.247}{4} \\\\ & = 3.812 \;\;\; or \;\;\; -1.312 \end{aligned}$

i.e. $\;$ $x = 3.81$ $\;$ or $\;$ $x = -1.31$ $\;\;\;$ [correct to two decimal places]

Statistics - Histogram

Draw a histogram for the following frequency distribution and find the mode from the graph.

Class Interval	$0 - 5$	$5 - 10$	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$
Frequency	$2$	$5$	$10$	$14$	$8$	$5$

In the histogram, the highest rectangle represents the maximum frequency (or modal class).

Inside this rectangle, draw lines $AC$ and $BD$ diagonally from the upper corners $A$ and $D$ of adjacent rectangles.

Let the point of intersection of $AC$ and $BD$ be $K$.

Draw $KL$ perpendicular to the horizontal axis.

The value of point $L$ on the horizontal axis represents the mode and the class interval in which point $L$ lies is the modal class.

From the figure, mode $= 17$

Trigonometry - Heights and Distances

A straight road leads to the foot of a tower $200 \; m$ high. From the top of the tower, the angles of depression of two cars standing on the road are observed to be $45^\circ$ and $60^\circ$ respectively. Find the distance between the two cars. Give your answer correct to two decimal places. Take $\sqrt{3} = 1.732$.

$OT$: Tower of height $200 \; m$

$C_1, \; C_2$: Positions of the two cars

$C_1C_2$: Distance between the two cars

In $\triangle OTC_1$,

$\dfrac{OT}{OC_1} = \tan 60^\circ = \sqrt{3}$

i.e. $\;$ $OC_1 = \dfrac{OT}{\sqrt{3}} = \dfrac{200}{\sqrt{3}} \; m$ $\;\;\; \cdots \; (1)$

In $\triangle OTC_2$,

$\dfrac{OT}{OC_2} = \tan 45^\circ = 1$

i.e. $\;$ $OC_2 = OT = 200 \; m$ $\;\;\; \cdots \; (2)$

Now, $\;$ $C_1 C_2 = OC_2 - OC1$

i.e. $\;$ $C_1 C_2 = 200 - \dfrac{200}{\sqrt{3}}$ $\;\;\;$ [in view of equations $(1)$ and $(2)$]

i.e. $\;$ $C_1 C_2 = 200 \left(\dfrac{\sqrt{3} - 1}{\sqrt{3}}\right) = 84.53 \; m$ $\;\;\;$ [correct to two decimal places]

Ratio and Proportion

If $x$, $y$ and $z$ are in continued proportion, prove that $\;$ $\dfrac{\left(x + y\right)^2}{\left(y + z\right)^2} = \dfrac{x}{z}$

$x, \; y, \; z$ are in continued proportion.

$\implies$ $\dfrac{x}{y} = \dfrac{y}{z} = k \;$ (say)

$\implies$ $y = z k$ $\;\;\; \cdots \; (1)$

and $\;$ $x = yk = zk^2$ $\;\;\; \cdots \; (2)$

$\begin{aligned} LHS & = \dfrac{\left(x + y\right)^2}{\left(y + z\right)^2} \\\\ & = \dfrac{\left(zk^2 + zk\right)^2}{\left(zk + z\right)^2} \;\;\; \left[\text{in view of equations (1) and (2)}\right] \\\\ & = \dfrac{\left[zk \left(k + 1\right)\right]^2}{\left[z \left(k + 1\right)\right]^2} \\\\ & = k^2 \;\;\; \cdots \; (3) \\\\ RHS & = \dfrac{x}{z} \\\\ & = \dfrac{zk^2}{z} = k^2 \;\;\; \cdots \; (4) \end{aligned}$

$\therefore \;$ From equations $(3)$ and $(4)$, $\;\;$ $LHS = RHS$

i.e. $\;$ $\dfrac{\left(x + y\right)^2}{\left(y + z\right)^2} = \dfrac{x}{z}$

Hence proved.