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Q.
The reflection of the point $(4,-13)$ about the line $5 x+y+6=0$ is
Straight Lines
Solution:
Let $(h, k)$ be the point of the given point $(4,-13)$ about the line $5 x+y+6=0$. The mid-point of the line segment joining points $(h, k)$ and $(4,-13)$ is given by
$\left(\frac{h+4}{2}, \frac{k-13}{2}\right)$
This point lies on the given line, so we have
$5\left(\frac{h+4}{2}\right)+\frac{k-13}{2}+6=0$
or $5 h+k+19=0 .....$(i)
Again, the slope of the line joining points $(h, k)$ and $(4,-13)$ is given by $\frac{k+13}{h-4}$. This line is perpendicular to the given line and hence $(-5)\left(\frac{k+13}{h-4}\right)=-1$
This gives $ 5 k+65=h-4$
or $ h-5 k-69=0 .....$(ii)
On solving Eqs. (i) and (ii), we get
$h=-1 \text { and } k=-14$
Thus, the point $(-1,-14)$ is the reflection of the given point.