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