Straight lines/Straight line graphs

Algebra

Questions marked Higher are aimed at the Higher tier.

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Find the gradient of the line through $(1, 3)$ and $(5, 11)$.

2 marks
Write the equation of the line with gradient $3$ that passes through $(0, 5)$.

2 marks
A line passes through $(2, 1)$ and $(2, 7)$. Write its equation, and say what happens to the gradient.

3 marks
Does $y = 2x - 4$ pass through $(3, 2)$? Does it pass through $(1, -1)$?

3 marks
Find the equation of the line through $(1, 4)$ and $(3, 10)$.

4 marks
Find the equation of the line perpendicular to $y = 2x + 1$ that passes through $(4, 3)$. Higher

4 marks
Line $L$ passes through $(0, -2)$ and $(4, 6)$. Find its equation, the point where it crosses the $x$-axis, and the midpoint of the two given points.

5 marks
Triangle $ABC$ has vertices $A(1, 1)$, $B(5, 1)$ and $C(5, 4)$, as shown. Find the equation of $AC$, the length of $BC$, the area of the triangle, and the equation of the line through $C$ parallel to $AB$.

Triangle ABC on a coordinate grid

6 marks

Mark scheme

Answer. $2$
- $\dfrac{11 - 3}{5 - 1} = 2$
Answer. $y = 3x + 5$
- The $y$-intercept is $5$.
- $y = 3x + 5$
Answer. The equation is $x = 2$, and the gradient is undefined.
- The $x$-coordinate does not change.
- The line is vertical, so it has no finite gradient.
Answer. Yes, and no
- $2 \times 3 - 4 = 2$, so $(3, 2)$ is on the line.
- $2 \times 1 - 4 = -2$, not $-1$.
Answer. $y = 3x + 1$
- Gradient $= \dfrac{6}{2} = 3$.
- $y - 4 = 3(x - 1)$, so $y = 3x + 1$.
Answer. $y = -\dfrac{1}{2}x + 5$
- The perpendicular gradient is $-\dfrac{1}{2}$.
- $y - 3 = -\dfrac{1}{2}(x - 4)$
- $y = -\dfrac{1}{2}x + 5$
Answer. $y = 2x - 2$, it crosses at $(1, 0)$, and the midpoint is $(2, 2)$.
- Gradient $= 2$, so $y = 2x - 2$.
- $0 = 2x - 2$ gives $(1, 0)$.
- Midpoint $= (2, 2)$.
Answer. $y - 1 = \dfrac{3}{4}(x - 1)$, $BC = 3$, area $6$, and $y = 4$
- Gradient of $AC = \dfrac{3}{4}$.
- $BC = 3$ and $AB = 4$, so the area is $\dfrac{1}{2} \times 4 \times 3 = 6$.
- $AB$ is horizontal, so the parallel line through $C$ is $y = 4$.