Mathematics - Straight Lines Question with Solution | TestHub

Let $\mathrm{Q}(a,b)$ be the reflection of $\mathrm{P}(4,-13)$ in the line $5x+y+6=0$ Then the mid-point $R\left(\frac{a+4}{2},\frac{b-13}{2}\right)$ lies on $\begin{array}{rl}& 5x+y+6=0\\ & \therefore 5\left(\frac{a+4}{2}\right)+\frac{b-13}{2}+6=0\\ & \Rightarrow 5a+b+19=0...(i)\end{array}$ Also $\mathrm{PQ}$ is perpendicular to $5x+y+6=0$ Therefore $\frac{b+13}{a-4}\times \left(-\frac{5}{1}\right)=-1$ $\Rightarrow a-5b-69=0...(ii)$ Solving (i) and (ii), we get $a=-1,b=-14$