Coordinate Geometry — Distance and Midpoint | Year 10 Maths Unit 2

Think First

Before we begin: Point A is at $(2, 3)$ and Point B is at $(6, 8)$ on a coordinate plane. Without using a formula, can you estimate how far apart they are? What about the point exactly halfway between them?

Come back to this after you have worked through the lesson.

The Big Idea

Coordinate geometry bridges algebra and geometry on the Cartesian plane. The distance formula measures how far apart two points are, using Pythagoras' theorem in disguise. The midpoint formula finds the exact centre point between two locations by simply averaging their coordinates. These two formulas are essential tools for navigation, mapping, and computer graphics.

Know

The distance formula and midpoint formula. How these formulas connect to Pythagoras' theorem and averaging.

Understand

Why the distance formula is derived from Pythagoras' theorem. Why the midpoint formula averages the coordinates.

Can Do

Calculate the distance between any two points. Find the midpoint of an interval joining two points.

Lesson Objectives

Apply the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ correctly.
Apply the midpoint formula $M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$ correctly.
Use these formulas in real-world contexts involving maps and navigation.

Key Vocabulary

Coordinate plane

A two-dimensional plane formed by the intersection of a horizontal x-axis and a vertical y-axis.

Interval

The line segment joining two points on the coordinate plane.

Distance formula

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, derived from Pythagoras' theorem.

Midpoint

The point that divides an interval into two equal parts.

Spot the Trap

Wrong: Forgetting to square both differences in the distance formula. $d = \sqrt{(x_2 - x_1) + (y_2 - y_1)}$.

Right: Square both differences before adding: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Wrong: Adding the coordinates for the midpoint instead of averaging. $M = (x_1 + x_2, y_1 + y_2)$.

Right: Divide by 2: $M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$.

The Distance Formula

+5 XP

The distance between two points $A(x_1, y_1)$ and $B(x_2, y_2)$ on a coordinate plane can be found using a direct application of Pythagoras' theorem. The horizontal distance is $|x_2 - x_1|$ and the vertical distance is $|y_2 - y_1|$. These form the two shorter sides of a right-angled triangle, with the interval $AB$ as the hypotenuse.

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

Pythagoras in disguise

Label your points

Always identify $(x_1, y_1)$ and $(x_2, y_2)$ before substituting. It prevents sign errors.

Square both differences

$(x_2 - x_1)^2$ not $(x_2 - x_1)$. Both differences must be squared before adding.

GPS uses this

Navigation systems use the distance formula (in 3D) to calculate distances between locations.

The Midpoint Formula

+5 XP

The midpoint of an interval is simply the average of the x-coordinates and the average of the y-coordinates. For points $A(x_1, y_1)$ and $B(x_2, y_2)$, the midpoint $M$ is exactly halfway between the two points in both the horizontal and vertical directions.

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

average of coordinates

Add then divide by 2

$(x_1 + x_2) / 2$ not $(x_1 + x_2)$. The midpoint is the average, not the sum.

Watch negative coordinates

$(-4 + 8) / 2 = 2$, not 6. Careful with signs when adding.

Verify with distance

Check that distance AM equals distance MB. If not, recalculate.

Finding an Endpoint

+5 XP

If you know the midpoint and one endpoint, you can work backwards to find the other endpoint. Rearrange the midpoint formula: if $M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$, then $x_2 = 2x_M - x_1$ and $y_2 = 2y_M - y_1$.

$x_2 = 2x_M - x_1$. $y_2 = 2y_M - y_1$.

work the formula backwards

Double then subtract

$x_2 = 2x_M - x_1$. Double the midpoint coordinate, then subtract the known endpoint.

Check your answer

Verify by calculating the midpoint of your two endpoints. It should match the given midpoint.

Think of it as reflection

The unknown endpoint is as far from the midpoint as the known endpoint, but in the opposite direction.

Real-World Applications

+5 XP

Coordinate geometry powers GPS navigation, computer graphics, and mapping systems. Every time your phone calculates the distance to a destination or a game renders a 3D scene, it uses these same formulas extended into three dimensions. Surveyors use coordinate geometry to measure land boundaries. Pilots use it for flight path calculations.

GPS navigation. Computer graphics. Surveying.

everywhere around you

Maps use coordinates

Every point on a digital map has coordinates. Distance and midpoint calculations happen constantly.

3D extension

In 3D graphics, a z-coordinate is added: $d = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}$.

Surveying

Land surveyors measure boundary distances and find central reference points using these exact formulas.

Worked Example 1 — Finding Distance

Q.Find the distance between $A(3, -2)$ and $B(7, 4)$.

Solution

Identify: $x_1 = 3$, $y_1 = -2$, $x_2 = 7$, $y_2 = 4$.

Substitute into distance formula: $d = \sqrt{(7 - 3)^2 + (4 - (-2))^2}$

$d = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \approx 7.21$

Answer: $\sqrt{52}$ or $2\sqrt{13}$ units (approximately 7.21 units).

Worked Example 2 — Finding the Midpoint

Q.Find the midpoint of the interval joining $P(-4, 6)$ and $Q(8, -2)$.

Solution

Method: $M = \left(\dfrac{-4 + 8}{2}, \dfrac{6 + (-2)}{2}\right)$

$M = \left(\dfrac{4}{2}, \dfrac{4}{2}\right) =$ $(2, 2)$

Check: Distance $PM = \sqrt{(2-(-4))^2 + (2-6)^2} = \sqrt{36+16} = \sqrt{52}$. Distance $MQ = \sqrt{(8-2)^2 + (-2-2)^2} = \sqrt{36+16} = \sqrt{52}$. Equal — confirmed!

Answer: $M = (2, 2)$

Worked Example 3 — Finding an Endpoint

Q.$M(5, -1)$ is the midpoint of $AB$. If $A$ is $(2, 3)$, find $B$.

Solution

Strategy: Use the midpoint formula backwards. $x_2 = 2x_M - x_1$ and $y_2 = 2y_M - y_1$.

$x_B = 2(5) - 2 =$ $8$

$y_B = 2(-1) - 3 =$ $-5$

Check: Midpoint of $(2, 3)$ and $(8, -5)$ is $\left(\dfrac{10}{2}, \dfrac{-2}{2}\right) = (5, -1)$ ✓

Answer: $B = (8, -5)$

Brain Trainer

4 quick-fire drills. Beat the clock.

Find the distance between $(0, 0)$ and $(3, 4)$.

Find the midpoint of $(2, 6)$ and $(8, -4)$.

(5, 1)

If midpoint is $(4, -3)$ and $A$ is $(1, 2)$, find $B$.

(7, -8)

Find the exact distance between $(1, -2)$ and $(4, 2)$.

Practice

5 MCQs and 3 short-answer questions. Target: 80% accuracy.

1.The distance between $(0, 0)$ and $(3, 4)$ is:

A3 B5 C7 D25

Correct! $d = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. This is the classic 3-4-5 right triangle.

Use the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Substitute $(0, 0)$ and $(3, 4)$.

2.The midpoint of $(2, 6)$ and $(8, -4)$ is:

A$(6, 1)$ B$(5, 1)$ C$(10, 2)$ D$(6, 2)$

Correct! $M = \left(\dfrac{2+8}{2}, \dfrac{6+(-4)}{2}\right) = \left(\dfrac{10}{2}, \dfrac{2}{2}\right) = (5, 1)$.

The midpoint formula averages the coordinates: $M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$. Add the coordinates, then divide by 2.

3.The distance formula is derived from:

ASimilar triangles BPythagoras' theorem CTrigonometric ratios DCoordinate addition

Correct! The horizontal difference $|x_2 - x_1|$ and vertical difference $|y_2 - y_1|$ form the two shorter sides of a right triangle. The distance is the hypotenuse: $c = \sqrt{a^2 + b^2}$.

Think about what geometric relationship connects the horizontal run, vertical rise, and straight-line distance between two points.

4.If $M(4, -3)$ is the midpoint of $AB$ and $A$ is $(1, 2)$, then $B$ is:

A$(5, -8)$ B$(7, -8)$ C$(2.5, -0.5)$ D$(5, -1)$

Correct! $x_B = 2(4) - 1 = 7$ and $y_B = 2(-3) - 2 = -8$. So $B = (7, -8)$.

Work the midpoint formula backwards: $x_2 = 2x_M - x_1$ and $y_2 = 2y_M - y_1$. Double the midpoint coordinate, then subtract the known endpoint.

5.The exact distance between $(1, -2)$ and $(4, 2)$ is:

A$3$ B$4$ C$5$ D$\sqrt{7}$

Correct! $d = \sqrt{(4-1)^2 + (2-(-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. Another 3-4-5 triangle in disguise!

Calculate $(x_2 - x_1)$ and $(y_2 - y_1)$, square both, add, then take the square root. Watch the double negative: $2 - (-2) = 4$.

6.Find the exact distance between $A(-5, 2)$ and $B(3, -4)$, and find the midpoint of $AB$.

Your answer:

Correct substitution into distance formula

Exact distance found (10 units)

Midpoint calculated correctly

Model Answer

Distance: $d = \sqrt{(3-(-5))^2 + (-4-2)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} =$ 10 units

Midpoint: $M = \left(\dfrac{-5+3}{2}, \dfrac{2+(-4)}{2}\right) = \left(\dfrac{-2}{2}, \dfrac{-2}{2}\right) =$ $(-1, -1)$

7.The points $P(2, 5)$, $Q(8, 5)$ and $R(5, 9)$ form a triangle. (a) Show that $PQ = 6$ units. (b) Calculate the lengths of $PR$ and $QR$. (c) What type of triangle is $PQR$? Justify your answer.

Your answer:

$PQ = 6$ shown correctly

$PR = 5$ and $QR = 5$ calculated

Triangle type identified with justification

Model Answer

(a) $PQ = \sqrt{(8-2)^2 + (5-5)^2} = \sqrt{36 + 0} = \sqrt{36} = 6$ ✓

(b) $PR = \sqrt{(5-2)^2 + (9-5)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

$QR = \sqrt{(5-8)^2 + (9-5)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

(c) Isosceles triangle because $PR = QR = 5$ (two equal sides).

8.A hiker walks from camp at $C(1, 2)$ to a lookout at $L(7, 10)$ on a coordinate map where each unit represents 1 km. (a) Calculate the direct distance from camp to lookout, correct to 1 decimal place. (b) Find the coordinates of the point halfway between camp and lookout. (c) The hiker then walks from the lookout to a waterfall at $W(11, 4)$. Calculate the total distance travelled (camp to lookout to waterfall).

Your answer:

Distance CL calculated correctly (1 mark)

Midpoint found correctly (1 mark)

LW calculated and total found (2 marks)

Correct units throughout (1 mark)

Model Answer

(a) $CL = \sqrt{(7-1)^2 + (10-2)^2} = \sqrt{36 + 64} = \sqrt{100} =$ 10.0 km

(b) Midpoint $= \left(\dfrac{1+7}{2}, \dfrac{2+10}{2}\right) =$ $(4, 6)$

(c) $LW = \sqrt{(11-7)^2 + (4-10)^2} = \sqrt{16 + 36} = \sqrt{52} \approx 7.2$ km

Total distance $= 10 + 7.2 =$ 17.2 km

Review

Consolidate and reflect before moving on.

Stretch Challenge

Prove that the triangle with vertices $A(1, 2)$, $B(5, 8)$ and $C(9, 2)$ is isosceles. Then find the exact length of the line of symmetry from the apex to the base. (Hint: find which two sides are equal, then find the midpoint of the base.)

Key Idea

Distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Midpoint formula: $M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$. Both come from basic geometric principles.

Common Trap

Forgetting to square both differences in the distance formula, or forgetting to divide by 2 in the midpoint formula. Label your points before substituting.

Connection

The distance formula is Pythagoras' theorem in coordinate form. The midpoint formula is averaging in two dimensions. Both extend naturally into 3D space.

Interactive: Distance and Midpoint Explorer — drag points on a coordinate plane and see the distance and midpoint update in real time.

Distance Calculator

Midpoint Master

Navigator

Daily Challenge

Find the distance and midpoint for $(-3, 5)$ and $(4, -1)$. Time yourself — can you do both in under 60 seconds?

I have completed this lesson and checked my answers.

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