The reflection of the point (4,-13) in the line 5 x+y+6=0 , is

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