Graphs 2

3 marks · Normal

Line $L$ passes through the points $P(-6, 8)$ and $Q(0, -4)$. Without finding its full equation, show that $L$ crosses the x-axis at $x = 2$.

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At $y=0$, $x=2$.

Common mistakes: $x=-2$; $x=4$

Show mark scheme and worked solution

Marking points

M1 · 1 mark · Works out the gradient using the two points.

M2 · 1 mark · Forms equation using point and gradient, sets $y=0$, solves for $x$.

A1 · 1 mark · Arrives at $x=2$ for the x-intercept.

Worked solution

Gradient: $m = (-4 - 8)/(0 - (-6)) = -12/6 = -2$.

$L: y = -2x + c$. Using $(0, -4)$: $-4 = -2\times 0 + c$, so $c = -4$.

At x-axis, $y=0$: $0 = -2x - 4 \implies 2x = -4 \implies x = -2$.

Correction: calculate $c$ again using $(0,-4)$: $-4 = -2\times0 + c \implies c = -4$.

At $y=0$: $0 = -2x -4 \implies 2x = -4 \implies x=-2$.