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Coordinate Geometry Problems

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AQA Level 2 Further Maths practice questions, each with a hint and a full worked solution.

Have a pen and paper handy, to work out your answers.

Question 12 marks

A is the point (-3, 5)

The midpoint of the line segment AB is M(4, -1)

Work out the coordinates of B.

Write your answer as (x, y)

Hint

The step from A to M is the same as the step from M to B.

Worked solution
  1. From A to M: x goes up by 4-(-3)=7 and y goes down by 5-(-1)=6
  2. Do the same step again from M to B
  3. x: \;4+7=11
  4. y: \;-1-6=-7
  5. (Check: \dfrac{-3+11}{2}=4 and \dfrac{5+(-7)}{2}=-1)
  6. B=(11, -7)

Question 22 marks

P is the point (-2, -3) and Q is the point (4, 1)

Work out the length of PQ.

Give your answer in the form a\sqrt{b} where a and b are integers and b is as small as possible.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Find the change in x and the change in y, then use Pythagoras; look for a square number factor at the end.

Worked solution
  1. Change in x: \;4-(-2)=6
  2. Change in y: \;1-(-3)=4
  3. Pythagoras: \;PQ^2=6^2+4^2=36+16=52
  4. PQ=\sqrt{52}=\sqrt{4}\times\sqrt{13}
  5. PQ=2\sqrt{13}

Question 3Challenge6 marks

A is the point (-4, 3) and B is the point (4, -1)

The point P lies on the line \;2x+y=17

P is the same distance from A as it is from B.

(a)

Work out the coordinates of P.

4 marks

Write your answer as (x, y)

(b)

Work out the area of triangle ABP.

2 marks

Hint

Points that are the same distance from A and B lie on the perpendicular bisector of AB; find its equation first.

Worked solution

Part (a)

  1. P is the same distance from A and B, so it lies on the perpendicular bisector of AB
  2. Midpoint of AB: \;M=\left(\dfrac{-4+4}{2}, \dfrac{3+(-1)}{2}\right)=(0, 1)
  3. Gradient of AB=\dfrac{-1-3}{4-(-4)}=\dfrac{-4}{8}=-\dfrac12
  4. Perpendicular gradient =2, so the perpendicular bisector is \;y=2x+1
  5. Substitute into 2x+y=17: \;2x+2x+1=17, so 4x=16 and x=4
  6. y=2\times4+1=9
  7. (Check: PA^2=8^2+6^2=100 and PB^2=0^2+10^2=100)
  8. P=(4, 9)

Part (b)

  1. Triangle ABP is isosceles (PA=PB), so PM is perpendicular to AB and is the height
  2. Base: \;AB=\sqrt{8^2+4^2}=\sqrt{80}=4\sqrt5
  3. Height: \;PM=\sqrt{4^2+8^2}=\sqrt{80}=4\sqrt5
  4. Area =\dfrac12\times4\sqrt5\times4\sqrt5=\dfrac12\times80
  5. Area =40 square units

More on this topic: Coordinate Geometry Problems worksheet with full solutions

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