Question

Let the point P(-8, 12) lies on the top of Qutub Minar, Delhi. Find the image of the point on the line $4x+7y+13=0.$

Answer 1

Let the image of the point P(-8, 12) in the line AB be Q(h, k), then, PQ is perpendicular bisected at R.

Then, coordinates of R are $(\frac{h-8}{2},\frac{k+12}{2}).$

Since, R lies on line $4x+7y+13=0,$ so it satisfies

$\frac{4(h-8)}{2}+\frac{7(k+12)}{2}+13=0$

$\Rightarrow 4h-32+7k+84+26=0$

$\Rightarrow 4h+7k+78=0...\left(i\right)$

Since, PQ! is perpendicular to AB.

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

$\Rightarrow -\frac{4}{7}\times \frac{k-12}{h+8}=-1$

$\Rightarrow 4k-48=7h+56$

$\Rightarrow 7h-4k+104=0...\left(ii\right)$

On solving Eqs. (i) and (ii), we get $h=-16,k=-2$

Hence, the image of (-8, 12) in the line $4x+7y+13=0$ is (-16, -2).