Question

The reflection of the point (4, –13) about the line 5x + y + 6 = 0 is
(a) (–1, –14)
(b) (3, 4)
(c) (0, 0)
(d) (1, 2)

Answer 1

Let A' (a, b) be the point of reflection of (4, −13) about the line 5x + y + 6 = 0
Here midpoint of AA' (4, −13) and (a, b) is given by $\left(\frac{4+a}{2},\frac{b-13}{2}\right)$
$$\begin{gathered} \mathrm{Since}\left(\frac{4+a}{2},\frac{b-13}{2}\right)\mathrm{lie}\mathrm{an}\left(5x+y+6\right)=0 \\ \mathrm{i}.\mathrm{e}.5\left(\frac{4+a}{2}\right)+\left(\frac{b-13}{2}\right)+6=0 \\ \mathrm{i}.\mathrm{e}.20+5a+b-13+12=0 \\ \mathrm{i}.\mathrm{e}.5a+b+19=0...\left(1\right) \\ \mathrm{Since}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{joining}\left(a,b\right)\mathrm{and}\left(4,-13\right)\mathrm{is}\mathrm{given}\mathrm{by}\frac{b+13}{a-4} \end{gathered}$$
Since line is perpendicular to the given line,
$$\begin{gathered} \left(-5\right)\left(\frac{b+13}{a-4}\right)=-1 \\ \mathrm{i}.\mathrm{e}.-5b-65=-a+4 \\ \mathrm{i}.\mathrm{e}.5b+65=a-4 \\ \mathrm{i}.\mathrm{e}.a-5b-69=0...\left(2\right) \\ \mathrm{Solving}\left(1\right)\mathrm{and}\left(2\right), \\ \frac{\begin{array}{c}\begin{array}{c}25a+5b+19\times 5=0\end{array}\\ a=5b-69=0\end{array}}{26a+26=0} \\ \mathrm{i}.\mathrm{e}.a=-1 \\ -5+b+19=0 \\ \mathrm{i}.\mathrm{e}.b=-14 \\ \mathrm{i}.\mathrm{e}.\left(a,b\right)=\left(-1,-14\right) \end{gathered}$$

Therefore reflection of (4, −13) about the line 5x + y + 6 = 0 is (−1, −14)

Hence, the correct answer is option A.