Coordinates and coordinate geometry
Algebra
- Crossover
- Non-calculator
- 8 questions
- 29 marks
Questions marked Higher are aimed at the Higher tier.
These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.
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Find the midpoint of $(2, 6)$ and $(8, 2)$.
2 marks
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The midpoint of $A(1, 4)$ and $B(p, 10)$ is $(4, 7)$. Find $p$.
2 marks
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Find the distance between $(1, 2)$ and $(4, 6)$.
3 marks
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For the line from $(0, 0)$ to $(6, 3)$, find the midpoint and the gradient.
3 marks
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Point $P$ divides the line from $(1, 2)$ to $(7, 8)$ in the ratio $1:2$. Find the coordinates of $P$.
4 marks
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The points are $A(2, 1)$, $B(8, 1)$ and $C(8, 5)$. What type of triangle is $ABC$, and what is its area?
4 marks
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For $A(0, 2)$, $B(6, 5)$ and $C(2, 8)$, find the gradient of $AB$, the length of $AB$, and the midpoint of $AC$. Higher
5 marks
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A rectangle has vertices $(1, 2)$, $(7, 2)$ and $(7, 6)$. Find the fourth vertex, the perimeter, the diagonal length, and the equation of the line through $(1, 2)$ and $(7, 6)$. Higher
6 marks
Mark scheme
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Answer. $(5, 4)$
- $\left(\dfrac{2 + 8}{2}, \dfrac{6 + 2}{2}\right)$
- $(5, 4)$
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Answer. $p = 7$
- $\dfrac{1 + p}{2} = 4$
- $p = 7$
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Answer. $5$
- Differences $3$ and $4$.
- $\sqrt{3^{2} + 4^{2}} = 5$
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Answer. Midpoint $(3, 1.5)$, gradient $\dfrac{1}{2}$
- Midpoint $= (3, 1.5)$.
- Gradient $= \dfrac{3}{6} = \dfrac{1}{2}$.
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Answer. $(3, 4)$
- $P$ is one third of the way from the first point to the second.
- $(1, 2) + \dfrac{1}{3}(6, 6) = (3, 4)$.
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Answer. Right-angled at $B$, area $12$
- $AB$ is horizontal and $BC$ is vertical, so the right angle is at $B$.
- Area $= \dfrac{1}{2} \times 6 \times 4 = 12$.
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Answer. Gradient $\dfrac{1}{2}$, length $3\sqrt{5}$, midpoint $(1, 5)$
- Gradient of $AB = \dfrac{3}{6} = \dfrac{1}{2}$.
- Length $= \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}$.
- Midpoint of $AC = (1, 5)$.
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Answer. Fourth vertex $(1, 6)$, perimeter $20$, diagonal $2\sqrt{13}$, line $y = \dfrac{2}{3}x + \dfrac{4}{3}$
- The fourth vertex is $(1, 6)$.
- Sides $6$ and $4$, so the perimeter is $20$.
- Diagonal $= \sqrt{36 + 16} = 2\sqrt{13}$.
- Gradient $\dfrac{2}{3}$ and $y - 2 = \dfrac{2}{3}(x - 1)$, so $y = \dfrac{2}{3}x + \dfrac{4}{3}$.