A line is drawn perpendicular to $5x = y + 7$. Find the equation of the line if the area of the triangle formed by this line with the coordinate axes is 10 square units.
Equation of the given line can be written as: $\hspace{1em}$ $y = 5x - 7$
$\therefore$ $\;$ Slope of the given line $= m = 5$
$\because$ $\;$ The required line is perpendicular to the given line,
slope of the required line $= m_1 = \dfrac{-1}{m} = \dfrac{-1}{5}$ $\;\;\; \cdots \; (1)$
Let AB be the required line where $A \left(0,a\right)$ and $B \left(b,0\right)$.
The equation of the required line can be written the form $y - y_1 = m_1 \left(x - x_1\right)$ as
$y - 0 = \dfrac{-1}{5} \left(x - b\right)$
i.e. $5y = -x + b$ $\implies$ $b = x + 5y$ $\;\;\; \cdots \; (2)$
The equation of the required line can also be written as
$y - a = \dfrac{-1}{5} \left(x - 0\right)$
i.e. $5 y - 5 a = - x$ $\implies$ $x + 5 y = 5a$ $\;\;\; \cdots \; (3)$
Given: The required line makes a triangle $\left(\triangle AOB\right)$ of area 10 square units with the coordinate axes.
Now, area of $\triangle AOB = \dfrac{1}{2}ab$
$\therefore$ $\;$ We have $\hspace{1em}$ $\dfrac{1}{2}ab = 10$ $\implies$ $a = \dfrac{20}{b}$ $\;\;\; \cdots \; (4)$
Substituting the value of $a$ from equation $(4)$ in equation $(3)$ gives
$x + 5y = 5 \times \dfrac{20}{b}$
$\implies$ $b = \dfrac{100}{x + 5y}$ $\;\;\; \cdots \; (5)$
$\therefore$ $\;$ We have from equations $(5)$ and $(2)$,
$x + 5y = \dfrac{100}{x + 5y}$
i.e. $\left(x + 5y\right)^2 = 100$
$\implies$ $x + 5 y = \pm 10$ $\;\;\; \cdots \; (6)$
Equation $(6)$ gives the equations of the required lines.