Coordinate Geometry: Question 1

Syllabus 1.3

Multiple choice AS 1 mark

Points C(4,1)C(-4, 1) and D(2,7)D(2, -7) lie on a coordinate grid.

What is the gradient of the line CDCD?

Choose an answer to check it, then compare with the worked solution below.

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Worked solution

Step 1: Recall the gradient formula

For two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), the gradient of the line through them is

m=y2y1x2x1.m = \frac{y_2 - y_1}{x_2 - x_1}.

Step 2: Substitute the coordinates

Take C(4,1)C(-4, 1) as (x1,y1)(x_1, y_1) and D(2,7)D(2, -7) as (x2,y2)(x_2, y_2):

m=712(4)=86=43.m = \frac{-7 - 1}{2 - (-4)} = \frac{-8}{6} = -\frac{4}{3}.

Checking the arithmetic a second time: the change in yy is 71=8-7 - 1 = -8, and the change in xx is 2(4)=2+4=62 - (-4) = 2 + 4 = 6, so m=8÷6=43m = -8 \div 6 = -\dfrac{4}{3}, confirming the same result.

Step 3: Sanity check

Moving from CC to DD, xx increases (from 4-4 to 22) while yy decreases (from 11 to 7-7), so the line falls from left to right. The gradient must be negative. This is consistent with m=43m = -\dfrac{4}{3}.

Why the other options are wrong

  • B (43)\left(\dfrac{4}{3}\right): correct size, but the negative sign has been lost. This would describe a line that rises from CC to DD, which contradicts the coordinates.
  • C (34)\left(-\dfrac{3}{4}\right): the gradient fraction has been inverted, using ΔxΔy\dfrac{\Delta x}{\Delta y} instead of ΔyΔx\dfrac{\Delta y}{\Delta x}.
  • D (34)\left(\dfrac{3}{4}\right): both the sign error and the inversion error have been made together.

Final answer

m=43(Option A)\boxed{m = -\frac{4}{3}} \quad \text{(Option A)}