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

Let line AB be $x+3y=7$ and point P be $(3,8)$.

Let $Q(h,k)$ be the image of point $P(3,8)$ in the line $x+3y=7$.

Since line $AB$ is a mirror,

1) Point $P$ and $Q$ are at equal distance from line $AB$, i.e., $PR=QR$, i.e., $R$ is the mid-point of $PQ.$

2) Image is formed perpendicular to mirror i.e., line PQ is perpendicular to line AB.

Since R is the midpoint of PQ.

Mid point of $PQ$ joining $(3,8)$ and $(h,k)$ is $(\frac{h+3}{2},\frac{k+8}{2})$

Coordinate of point R = $(\frac{h+3}{2},\frac{k+8}{2})$

Since point R lies on the line AB.

Therefore,

$\left(\frac{3+h}{2}\right)+3\left(\frac{8+k}{2}\right)=7$

$h+3k=-13$ ....(1)

Also, $PQ$ is perpendicular to $AB.$

Therefore,

Slope of $PQ$ $\times$ Slope of $AB=-1$

Since, slope of $AB=$ $-\frac{1}{3}$

Therefore, slope of $PQ=$ $3$

Now, PQ is line joining $P(3,8)$ and $Q(h,k).$

Slope of PQ = $3=\frac{k-8}{h-3}$

$3h-k=1$ ........(2)

Solving equation 1 and 2, we get,

$h=-1$ and $k=-4$

Hence, image is $Q(-1,-4)$.