Find the image of the point $\left(-2,3\right)$ about the line $x + 2y - 9 = 0$
Let $P \left(-2.3\right)$ be the given point.
Perpendicular distance of point P from the given line $x + 2y - 9 = 0$ is
$d = \left|\dfrac{1 \times \left(-2\right) + 2 \times 3 - 9}{\sqrt{\left(1\right)^2 + \left(2\right)^2}}\right|$
i.e. $d = \left|\dfrac{-2 + 6 - 9}{\sqrt{5}}\right| = \sqrt{5}$ units $\;\;\; \cdots \; (1)$
Let $Q \left(a, b\right)$ be the image of the point $P$ in the given line.
Then, perpendicular distance of $Q$ from the given line is $= d = \sqrt{5}$ units
i.e. $\left|\dfrac{1 \times a + 2 \times b - 9}{\sqrt{\left(1\right)^2 + \left(2\right)^2}}\right| = \sqrt{5}$
i.e. $\dfrac{a + 2b - 9}{\sqrt{5}} = \sqrt{5}$
i.e. $a + 2b - 9 = 5$ $\implies$ $a = 14 - 2b$ $\;\;\; \cdots \; (2)$
Now, distance $PQ = \sqrt{5} + \sqrt{5} = 2 \sqrt{5}$ units
Also, $PQ = \sqrt{\left(a + 2\right)^2 + \left(b - 3\right)^2}$
$\therefore$ $\;$ We have $\sqrt{\left(a + 2\right)^2 + \left(b - 3\right)^2} = 2 \sqrt{5}$
i.e. $\left(a + 2\right)^2 + \left(b - 3\right)^2 = 20$ $\;\;\; \cdots \; (3)$
Substituting the value of $a$ from equation $(2)$ in equation $(3)$ gives
$\left(14 - 2b + 2\right)^2 + \left(b - 3\right)^2 = 20$
i.e. $\left(16 - 2b\right)^2 + \left(b - 3\right)^2 = 20$
i.e. $256 - 64 b + 4 b^2 + b^2 - 6b + 9 = 20$
i.e. $5b^2 - 70b + 245 = 0$
i.e. $b^2 - 14 b + 49 = 0$
i.e. $\left(b - 7\right)^2 = 0$ $\implies$ $b - 7 = 0$ $\implies$ $b = 7$ $\;\;\; \cdots \; (4)$
Substituting the value of $b$ from equation $(4)$ in equation $(2)$ gives
$a = 14 - 2 \times 7 = 0$
$\therefore$ $\;$ The image of the point $\left(-2,3\right)$ about the given line is $\left(0,7\right)$