The reflection of the point (4, -13) in the l | vedclass

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(A) Let $Q(a, b)$ be the reflection of $P(4, -13)$ in the line $5x + y + 6 = 0$.The mid-point $R\left(\frac{a + 4}{2}, \frac{b - 13}{2}\right)$ lies o

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$(-1, -14)$

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(A) Let $Q(a, b)$ be the reflection of $P(4, -13)$ in the line $5x + y + 6 = 0$.The mid-point $R\left(\frac{a + 4}{2}, \frac{b - 13}{2}\right)$ lies on the line $5x + y + 6 = 0$.Substituting the coordinates of $R$ into the line equation:$5\left(\frac{a + 4}{2}\right) + \left(\frac{b - 13}{2}\right) + 6 = 0$$5(a + 4) + (b - 13) + 12 = 0$$5a + 20 + b - 13 + 12 = 0$$5a + b + 19 = 0$ ......$(i)$Also,the line segment $PQ$ is perpendicular to the line $5x + y + 6 = 0$.Since the slope of the line is $-5$,the slope of $PQ$ must be $\frac{1}{5}$.$\frac{b - (-13)}{a - 4} = \frac{1}{5}$$5(b + 13) = a - 4$$5b + 65 = a - 4$$a - 5b - 69 = 0$ ......$(ii)$Solving equations $(i)$ and $(ii)$:From $(i)$,$b = -5a - 19$.Substituting into $(ii)$:$a - 5(-5a - 19) - 69 = 0$$a + 25a + 95 - 69 = 0$$26a + 26 = 0$$a = -1$Substituting $a = -1$ into $b = -5a - 19$:$b = -5(-1) - 19 = 5 - 19 = -14$.Thus,the reflection is $(-1, -14)$.

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

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