Question

If $(h, k)$ is the image of the point $(2, -3)$ with respect to the line $5x - 3y = 2$, then $h + k =$

• A. $-3$
• B. $\frac{3}{34}$
• C. $-\frac{1}{34}$
• D. $5$

Hint:

For reflection, the midpoint of the point and its image lies on the mirror line, and the segment joining them is perpendicular to the line. Solve the resulting system of equations.

Answer 1

The Correct Option is A

The image $(h, k)$ of point $P(2, -3)$ with respect to the line $5x - 3y = 2$ satisfies: 1. The midpoint of $PQ$ lies on the line. 2. Line $PQ$ is perpendicular to the given line. Midpoint condition: Midpoint $M\left( \frac{2 + h}{2}, \frac{-3 + k}{2} \right)$ lies on $5x - 3y = 2$: \[ 5 \left( \frac{2 + h}{2} \right) - 3 \left( \frac{-3 + k}{2} \right) = 2 \] \[ 5 (2 + h) - 3 (-3 + k) = 4 \implies 10 + 5h + 9 - 3k = 4 \implies 5h - 3k = -15 \] Perpendicularity condition: Slope of the line $5x - 3y = 2$ is $\frac{5}{3}$. Slope of $PQ$: \[ \frac{k - (-3)}{h - 2} = \frac{k + 3}{h - 2} \] Perpendicularity gives: \[ \left( \frac{k + 3}{h - 2} \right) \cdot \frac{5}{3} = -1 \implies 5 (k + 3) = -3 (h - 2) \] \[ 5k + 15 = -3h + 6 \implies 3h + 5k = -9 \] Solve: \[ 5h - 3k = -15 \quad (1) \] \[ 3h + 5k = -9 \quad (2) \] Multiply (1) by 5 and (2) by 3: \[ 25h - 15k = -75 \] \[ 9h + 15k = -27 \] Add: \[ 34h = -102 \implies h = -3 \] Substitute $h = -3$ into (2): \[ 3(-3) + 5k = -9 \implies -9 + 5k = -9 \implies k = 0 \] \[ h + k = -3 + 0 = -3 \] Option (1) is correct. Options (2), (3), and (4) do not match.