Alliance Form 3 P1 Q20 — Coordinate Geometry: Equations of Lines

KCSE (Form 3) Form 3 Geometry

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The Question

“Line L passes through the points (-2, 3) and (-1, 6). Line P is perpendicular to L at the point (-1, 6). Line Q is parallel to L and passes through the point (1, 2). Part A: find the equation of line L. Part B: find the equation of line P. Part C: find the equation of line Q and its x and y intercepts. Part D: find the point where lines P and Q intersect.”

1

Find the gradient and equation of L

The gradient of a line is the change in the y-values divided by the change in the x-values between two known points. Once you have the gradient, use one of the points to build the equation with the point-gradient form, then rearrange it into the y = mx + c style so it is easy to read off.

mL=6−3−1−(−2)=31=3m_{L} = \frac{6 - 3}{-1 - (-2)} = \frac{3}{1} = 3
y−6=3(x+1)y - 6 = 3(x + 1)
y=3x+9(line L)y = 3x + 9 \quad \text{(line L)}
2

Find the equation of the perpendicular line P

For two lines to be perpendicular, the product of their gradients must equal minus one. So flip the gradient of L and change its sign to get the gradient of P. Line P passes through the point of contact, so substitute that point into the point-gradient form and tidy up.

mP×3=−1⇒mP=−13m_{P} \times 3 = -1 \Rightarrow m_{P} = -\frac{1}{3}
y−6=−13(x+1)y - 6 = -\frac{1}{3}(x + 1)
x+3y=17(line P)x + 3y = 17 \quad \text{(line P)}
3

Find the equation of the parallel line Q and its intercepts

Parallel lines have exactly the same gradient, so Q also has gradient 3. Use the point it passes through to form its equation. To find the intercepts, set y to zero to get where it crosses the x-axis, and set x to zero to get where it crosses the y-axis.

y−2=3(x−1)⇒y=3x−1(line Q)y - 2 = 3(x - 1) \Rightarrow y = 3x - 1 \quad \text{(line Q)}
y=0⇒x=13(x-intercept)y = 0 \Rightarrow x = \frac{1}{3} \quad \text{(x-intercept)}
x=0⇒y=−1(y-intercept)x = 0 \Rightarrow y = -1 \quad \text{(y-intercept)}
4

Find where P and Q intersect

The point of intersection is where both equations are true at once, so solve them simultaneously. Substitute the expression for y from line Q into the equation of line P, solve for x, then put that value back into either equation to get y.

x+3(3x−1)=17x + 3(3x - 1) = 17
10x−3=17⇒10x=20⇒x=210x - 3 = 17 \Rightarrow 10x = 20 \Rightarrow x = 2
y=3(2)−1=5y = 3(2) - 1 = 5

Final Result

Line L is y = 3x + 9. Line P is x + 3y = 17. Line Q is y = 3x - 1, crossing the x-axis at x = 1/3 and the y-axis at y = -1. Lines P and Q intersect at the point (2, 5).

Why this method works

The whole question hangs on three facts about gradients: the gradient measures steepness as rise over run, perpendicular gradients multiply to minus one, and parallel lines share a gradient. Once the gradient of L is known, every other line is built from it. Finding the intersection works because a point that lies on two lines must satisfy both equations, so solving them together pins down the single x and y that fit both — here the coordinates (2, 5).

✓

Substitute x = 2, y = 5 into line P: 2 + 3(5) = 2 + 15 = 17, and into line Q: 3(2) - 1 = 5. Both hold, so (2, 5) is correct.