Straight lines
Coordinate geometry
- AS
- Non-calculator
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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Find the gradient of the line through $A(1, 4)$ and $B(5, -2)$, and the equation of the line $AB$.
3 marks
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Show that the lines $2x + 3y - 6 = 0$ and $3x - 2y + 1 = 0$ are perpendicular.
3 marks
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The points $A$ and $B$ are $(2, -1)$ and $(8, 5)$. Find the midpoint of $AB$ and the distance $AB$.
4 marks
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Find the equation of the line through $(3, 4)$ that is parallel to $y = 2x - 5$.
4 marks
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The points $P$ and $Q$ are $(1, 2)$ and $(4, 6)$. Find the equation of the perpendicular bisector of $PQ$.
5 marks
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The line $L$ passes through $(0, 3)$ and is perpendicular to $4x - 2y + 1 = 0$. Find the equation of $L$.
5 marks
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The points are $A(1, 1)$, $B(4, 2)$ and $C(3, 5)$. Show that $AB$ is perpendicular to $BC$, and find the area of triangle $ABC$.
6 marks
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The distance from $A(2, 1)$ to $B(8, k)$ is $10$. Find the possible values of $k$, and the equation of the perpendicular bisector of $AB$ in each case.
7 marks
Mark scheme
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Answer. The gradient is $-\dfrac{3}{2}$ and the equation is $y - 4 = -\dfrac{3}{2}(x - 1)$.
- Gradient $= \dfrac{-2 - 4}{5 - 1} = -\dfrac{3}{2}$.
- Use $y - 4 = -\dfrac{3}{2}(x - 1)$.
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Answer. The gradients are $-\dfrac{2}{3}$ and $\dfrac{3}{2}$, and their product is $-1$.
- Rewrite as $y = -\dfrac{2}{3}x + 2$ and $y = \dfrac{3}{2}x + \dfrac{1}{2}$.
- The product of the gradients is $-1$, so the lines are perpendicular.
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Answer. The midpoint is $(5, 2)$ and $AB = 6\sqrt{2}$.
- Midpoint $= \left(\dfrac{2 + 8}{2}, \dfrac{-1 + 5}{2}\right) = (5, 2)$.
- $AB = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}$.
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Answer. $y = 2x - 2$
- A parallel line has gradient $2$.
- $y - 4 = 2(x - 3)$.
- $y = 2x - 2$.
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Answer. $y - 4 = -\dfrac{3}{4}\left(x - \dfrac{5}{2}\right)$
- Midpoint $= \left(\dfrac{5}{2}, 4\right)$.
- Gradient of $PQ$ is $\dfrac{4}{3}$, so the perpendicular gradient is $-\dfrac{3}{4}$.
- $y - 4 = -\dfrac{3}{4}\left(x - \dfrac{5}{2}\right)$.
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Answer. $y = -\dfrac{1}{2}x + 3$
- The given line has gradient $2$.
- The perpendicular gradient is $-\dfrac{1}{2}$.
- $L$ has equation $y = -\dfrac{1}{2}x + 3$.
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Answer. $AB$ is perpendicular to $BC$, and the area is $5$.
- $\overrightarrow{AB} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$ and $\overrightarrow{BC} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}$.
- The scalar product is $-3 + 3 = 0$, so the sides are perpendicular.
- $AB = BC = \sqrt{10}$.
- Area $= \dfrac{1}{2} \times \sqrt{10} \times \sqrt{10} = 5$.
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Answer. $k = 9$ or $k = -7$. The bisectors are $y - 5 = -\dfrac{3}{4}(x - 5)$ and $y + 3 = \dfrac{3}{4}(x - 5)$.
- $6^2 + (k - 1)^2 = 100$, so $(k - 1)^2 = 64$.
- $k - 1 = 8$ or $k - 1 = -8$, giving $k = 9$ or $k = -7$.
- If $k = 9$, the midpoint is $(5, 5)$ and the perpendicular gradient is $-\dfrac{3}{4}$.
- If $k = -7$, the midpoint is $(5, -3)$ and the perpendicular gradient is $\dfrac{3}{4}$.