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9709 P12 - Jun 2023 - Q7
101
Find the coordinates of the reflection of the point
(-1, 3) across the line 3y + 2x = 33.
Solution
1. The equation of the line is given as \(3y + 2x = 33\).
Rearranging gives \(y = -\frac{2}{3}x + 11\).
2. The gradient of the line is \(-\frac{2}{3}\).
The gradient of the perpendicular line is the negative reciprocal,
\(\frac{3}{2}\).
3. The equation of the line perpendicular to
3y + 2x = 33 and passing through (-1, 3) is
\(y - 3 = \frac{3}{2}(x + 1)\).
4. Simplifying gives the equation
\(y = \frac{3}{2}x + \frac{9}{2}\).
5. Solving the system of equations
\(3y + 2x = 33\) and
\(y = \frac{3}{2}x + \frac{9}{2}\) gives the intersection point
(3, 9).
6. The reflection of (-1, 3) across the line is found by
using the midpoint formula. The midpoint of (-1, 3) and
the reflection point R is (3, 9). Solving gives
R = (7, 15).