Question

Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

Answer 1

The equation of the given line is
$x + 3y = 7 … (1)$

Let point B (a, b) be the image of point A (3, 8).
Accordingly, line (1) is the perpendicular bisector of AB.

Slope of$AB =\frac{b-8}{a-3}$, while the slope of the line $(l)=-\frac{1}{3}$

Since line (1) is perpendicular to AB,
$\left(\frac{b-8}{a-3}\right)\times\left(\frac{-1}{3}\right)=-1$

$⇒\frac{ b-8}{3a-9}=1$

$⇒ b-8=3a-9$
$⇒ 3a-b=1 ....(2)$

Mid-Point of $AB =\left(\frac{a+3}{2},\frac{b+8}{2}\right)$
The mid-point of line segment AB will also satisfy line (1).
Hence, from equation (1), we have

$\left(\frac{a+3}{2}\right)+3\left(\frac{b+8}{2}\right)=7$

$⇒ a+3+3b+24=14$

$⇒ a+3b=-13 ......(3)$
On solving equations (2) and (3), we obtain a = -1 and b = -4.
Thus, the image of the given point with respect to the given line is (-1, -4).