Perpendicular Lines and Locus: Coordinate Geometry Question

Grades 10-11
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Question: In the figure, the straight line L1: 3x – 2y + 6 = 0 and the straight line L2 are perpendicular to each other and intersect at A. It is given that L1 cuts the y-axis at C and L2 passes through the point (6, 7). (a) Find the equation of L2. (3 marks) (b) Q is a moving point in the coordinate plane such that AQ = CQ. Denote the locus of Q by . (i) Describe the geometric relationship between  and L2. Explain your answer. (ii) Find the equation of .
To solve this problem, we will look at the relationships between lines and points in the coordinate plane.

(a) Finding the Equation of L_2

First, we find the slope of L_1. By rewriting 3x - 2y + 6 = 0 in the slope-intercept form y = mx + c: 2y = 3x + 6 y = \frac{3}{2}x + 3 The slope of L_1 is m_1 = \frac{3}{2}.
Since L_1 and L_2 are perpendicular, the product of their slopes is -1. Let m_2 be the slope of L_2:
m_1 \cdot m_2 = -1
\frac{3}{2} \cdot m_2 = -1 \implies m_2 = -\frac{2}{3}
L_2 passes through the point (6, 7). Using the point-slope form:
y - 7 = -\frac{2}{3}(x - 6)
3(y - 7) = -2(x - 6)
3y - 21 = -2x + 12
2x + 3y - 33 = 0

(b) The Locus \Gamma

(i) Geometric Relationship between \Gamma and L_2

The moving point Q satisfies AQ = CQ. In geometry, the locus of points equidistant from two fixed points A and C is the perpendicular bisector of the segment connecting those points.
  1. Points A and C both lie on line L_1 (since C is the y-intercept of L_1 and A is the intersection of L_1 and L_2).
  2. Therefore, \Gamma is the perpendicular bisector of the segment AC which lies on L_1.
  3. Since \Gamma \perp L_1 and we are given L_2 \perp L_1, both lines are perpendicular to the same line.
  4. Thus, \Gamma is parallel to L_2.

(ii) Finding the Equation of \Gamma

First, we find the coordinates of C and A:
  • For C (y-intercept of L_1): set x = 0 in 3x - 2y + 6 = 0 \implies y = 3. So C = (0, 3).
  • For A (intersection of L_1 and L_2): Solve 3x - 2y = -6 and 2x + 3y = 33. Solving this system gives A = (\frac{48}{13}, \frac{111}{13}).
The midpoint M of AC is:
M = \left( \frac{0 + \frac{48}{13}}{2}, \frac{3 + \frac{111}{13}}{2} \right) = \left( \frac{24}{13}, \frac{75}{13} \right)
\Gamma has the same slope as L_2, which is m = -\frac{2}{3}, and passes through M:
y - \frac{75}{13} = -\frac{2}{3}\left(x - \frac{24}{13}\right)
Multiplying by 3 \times 13:
39y - 225 = -2(13x - 24)
39y - 225 = -26x + 48
26x + 39y - 273 = 0
Dividing the equation by 13:
2x + 3y - 21 = 0
The provided diagram visualizes L_1 (blue) and L_2 (red) intersecting at A. The locus \Gamma (green) is shown as the perpendicular bisector of the segment AC, remaining parallel to L_2.
coordinate-geometrylociperpendicular-lines