KCSE 2025 Maths P1 Q19 — Coordinate Geometry (Parallel & Perpendicular Lines)

KCSE 2025 Form 4 Geometry

Published

The Question

“Two lines L1 and L2 are parallel, and line L3 is perpendicular to both of them. Line L1 passes through the points A(-3, 6) and B(3, 9). (a)(i) Find the gradient of line L1. (a)(ii) Line L2 crosses the x-axis at C(-5, 0); determine its equation in the form y = mx + c. (b) Line L3 crosses L1 at A(-3, 6) and crosses L2 at point P. (b)(i) Determine the equation of L3 in the form y = mx + c. (b)(ii) Determine the coordinates of point P.”

1

Find the gradient of L1

The gradient measures how steeply a line rises. It is the change in the y-values divided by the change in the x-values between two points on the line. Using A(-3, 6) and B(3, 9), subtract the coordinates carefully, remembering that subtracting a negative x-value adds to the denominator.

m1=9−63−(−3)=36=12m_{1} = \frac{9 - 6}{3 - (-3)} = \frac{3}{6} = \frac{1}{2}
2

Write the equation of L2

Because L2 is parallel to L1, the two lines never meet and must have exactly the same gradient, one half. L2 passes through C(-5, 0), so substitute those coordinates into y = mx + c to solve for the y-intercept c, then write the full equation.

0=12(−5)+c⇒c=520 = \frac{1}{2}(-5) + c \Rightarrow c = \frac{5}{2}
L2:  y=12x+52L_{2}: \; y = \frac{1}{2}x + \frac{5}{2}
3

Write the equation of L3

L3 is perpendicular to L1, so its gradient is the negative reciprocal of one half. Flipping the fraction gives two and changing the sign gives negative two. L3 passes through A(-3, 6), so use the point-gradient form and simplify. Notice the constant terms cancel to give a clean equation through the origin.

m3=−1m1=−112=−2m_{3} = -\frac{1}{m_{1}} = -\frac{1}{\tfrac{1}{2}} = -2
y−6=−2(x−(−3))=−2(x+3)y - 6 = -2\big(x - (-3)\big) = -2(x + 3)
y−6=−2x−6⇒y=−2xy - 6 = -2x - 6 \Rightarrow y = -2x
4

Find point P where L3 meets L2

P is the point where lines L3 and L2 cross, so at P both equations give the same y for the same x. Set the two expressions for y equal, collect the x-terms on one side, and solve for x. Then substitute that x back into the simpler equation y = -2x to get the matching y-value.

−2x=12x+52-2x = \frac{1}{2}x + \frac{5}{2}
−52x=52⇒x=−1-\frac{5}{2}x = \frac{5}{2} \Rightarrow x = -1
y=−2(−1)=2y = -2(-1) = 2

Final Result

The gradient of L1 is one half. The equation of L2 is y = (1/2)x + 5/2. The equation of L3 is y = -2x. The lines L3 and L2 meet at point P(-1, 2).

Why this method works

Parallel lines share a gradient because they rise at the same rate and never converge, so copying L1's gradient onto L2 is valid. Perpendicular lines meet at a right angle, and this forces their gradients to multiply to negative one, which is why the negative reciprocal of one half is negative two. Finding an intersection works because a point that lies on two lines must satisfy both their equations at once; setting the equations equal locates the single x-value common to both, and substituting back recovers the shared y-value.

✓

Substitute P(-1, 2) into L2: (1/2)(-1) + 5/2 = -1/2 + 5/2 = 2, which matches, so P lies on L2 as required.