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Coordinate Geometry Problems

How to use distances, midpoints, ratios and gradients to solve coordinate geometry problems and justify geometric facts.

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Coordinate geometry turns a diagram into calculations. Use differences for distances and gradients, averages for midpoints, and a fraction of the displacement for a ratio.

For A(x_1,y_1) and B(x_2,y_2):

AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)

The distance formula is Pythagoras’ theorem applied to the horizontal and vertical changes. A distance is non-negative; its squared differences cannot cancel each other.

Worked example 1

For A(-4,1) and B(2,9), find the distance AB and the midpoint of AB.

  1. The horizontal change is 2-(-4)=6 and the vertical change is 9-1=8

  2. Your turn. Work out \sqrt{6^2+8^2}

    AB=\sqrt{36+64}=\sqrt{100}=10

  3. The midpoint is \left(\frac{-4+2}{2},\frac{1+9}{2}\right)=(-1,5)
  4. Answer: AB=10 and the midpoint is (-1,5)

Worked example 2

A=(-1,5). The midpoint of AB is M=(3,-2). Find B.

  1. Write B=(u,v). The midpoint equations are \frac{-1+u}{2}=3 and \frac{5+v}{2}=-2
  2. From the first equation, -1+u=6, so u=7

  3. Your turn. Solve 5+v=-4 for v

    v=-9

  4. Answer: B=(7,-9). Check by averaging the two endpoints: the result is (3,-2)

Dividing a segment in a ratio

If AP:PB=m:n and P lies between A and B, travel \frac{m}{m+n} of the way from A to B. Apply the same fraction to both coordinate changes.

The fraction is measured from the endpoint named first. For AP:PB=2:3, start at A and use \frac25, not \frac23

Explore a point on a segment

Try it: use 5 parts out of 10. This gives the midpoint. Compare the changes in both coordinates as you move the point.

Worked example 3

A=(-3,2) and B=(7,7). The point P lies on AB with AP:PB=2:3. Find P.

  1. The displacement from A to B is (10,5). There are 2+3=5 equal parts.
  2. Two fifths of the displacement is \left(\frac25\times10,\frac25\times5\right)=(4,2)

  3. Your turn. Add (4,2) to the coordinates of A. What are the coordinates of P?

    Write your answer as (x, y)

    P=(-3+4,2+2)=(1,4)

  4. Answer: P=(1,4). The remaining displacement is (6,3), which is three parts.

Points beyond the segment

Read the order of the points carefully. If A, B, C are in that order on a straight line, then C is beyond B. You must extend the displacement rather than choose a point between A and B

Worked example 4

A(-2,1), B(4,4) and C lie in that order on a straight line. Given AB:BC=3:2, find C.

  1. The displacement from A to B is (6,3). This represents three parts.
  2. One part is (2,1), so the displacement from B to C is (4,2)

  3. Your turn. Find C by starting at B=(4,4) and adding (4,2)

    Write your answer as (x, y)

    C=(4+4,4+2)=(8,6)

  4. Answer: C=(8,6). It lies beyond B, as required.

Perpendicular bisectors

The perpendicular bisector of AB passes through its midpoint and is at right angles to AB. Find the midpoint, find the perpendicular gradient, then form a line equation.

Every point on this line is equidistant from A and B. If AB is horizontal, its perpendicular bisector is vertical, and conversely.

Worked example 5

Find the perpendicular bisector of the segment joining A(-2,5) and B(6,1).

  1. Your turn. Find the midpoint of AB

    Write your answer as (x, y)

    The midpoint is \left(\frac{-2+6}{2},\frac{5+1}{2}\right)=(2,3)

  2. The gradient of AB is \frac{1-5}{6-(-2)}=-\frac12, so the perpendicular gradient is 2
  3. Through (2,3), the line is y-3=2(x-2)
  4. Answer: y=2x-1. It passes through the midpoint and has gradient perpendicular to AB

Proving geometric facts

For collinear points, show the relevant gradients are equal, or show that all the points satisfy the same line equation. For a right angle, show the adjacent gradients multiply to -1, or use Pythagoras on the three squared side lengths.

Name the sides and state your conclusion. A diagram that looks right-angled is not a proof.

Worked example 6

A=(-1,2), B=(3,4) and C=(6,-2). Prove that ABC is right-angled and find its area.

  1. The gradient of AB is \frac{4-2}{3-(-1)}=\frac12. The gradient of BC is \frac{-2-4}{6-3}=-2
  2. \frac12\times(-2)=-1, so AB and BC are perpendicular. The right angle is at B
  3. AB=\sqrt{4^2+2^2}=2\sqrt5 and BC=\sqrt{3^2+(-6)^2}=3\sqrt5. These are the perpendicular base and height.

  4. Your turn. Work out \frac12(2\sqrt5)(3\sqrt5)

    The area is \frac12\times6\times5=15 square units.

  5. Answer: the triangle is right-angled at B, and its area is 15 square units.

Common mistakes

  • Subtracting coordinates for a midpoint. Average each pair of coordinates instead.
  • Forgetting to square both coordinate changes. Square both changes, add them, then take the square root.
  • Using the wrong fraction for a ratio. For AP:PB=2:3, move \frac25 of the way from A to B.
  • Starting from the wrong endpoint. Check the ratio’s order and whether the point is inside or beyond the segment.
  • Finding only a perpendicular line. A perpendicular bisector must also pass through the midpoint.
  • Using a sloping side as a height. Triangle area needs a perpendicular base and height.

Now try it: Coordinate Geometry Problems practice questions

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