KJSEA 2025 Maths Q17 — Gradient of a Perpendicular Line

KJSEA 2025 Grade 9 Geometry

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The Question

“The equation of line L1 is 2y = x + 3. A line L2 is perpendicular to L1. What is the gradient of L2?”

1

Rewrite L1 in gradient form

To read off the gradient directly, put the equation into the standard form y = mx + c, where m is the gradient. The given equation has a 2 in front of y, so divide every term on both sides by 2 to make y stand alone.

2y=x+32y = x + 3
y=12x+32y = \frac{1}{2}x + \frac{3}{2}
2

Identify the gradient of L1

In the form y = mx + c, the number multiplying x is the gradient. Comparing the rearranged equation with this form shows the gradient of L1 is one half.

m1=12m_{1} = \frac{1}{2}
3

Apply the perpendicular gradient rule

When two lines are perpendicular, the product of their gradients is minus one. This means the gradient of L2 is the negative reciprocal of the gradient of L1: flip the fraction over and change its sign.

m1×m2=−1m_{1} \times m_{2} = -1
m2=−1m1=−112m_{2} = -\frac{1}{m_{1}} = -\frac{1}{\frac{1}{2}}
4

Simplify to find the gradient of L2

Flipping one half gives two, and the negative sign makes it minus two. That is the gradient of the perpendicular line L2.

m2=−2m_{2} = -2

Final Result

The gradient of L2 is -2 (letter C).

Why this method works

The rule works because rotating a line by 90 degrees swaps its rise and run and reverses one of their signs, so a slope of a/b becomes -b/a. That is exactly the negative reciprocal. Multiplying a gradient by its negative reciprocal always gives minus one, which is why perpendicular gradients satisfy that product. Here one half becomes minus two, confirming L1 and L2 cross at a right angle.

✓

Check the product: (1/2) times (-2) = -1, which confirms the two lines are perpendicular.