Calculate the distance between two points, the midpoint of a segment, and the gradient, and apply them to geometric problems

What this dot point is asking

SEAB wants you to find the distance between two points, the midpoint of the segment joining them, and the gradient of that segment, and to apply these in geometric problems such as showing a shape's properties. These three formulas turn coordinates into lengths, centres and slopes.

The answer

The distance between two points

The distance comes from Pythagoras theorem applied to the horizontal and vertical gaps:

Because the differences are squared, the order of subtraction does not matter for distance.

The midpoint

The midpoint of the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is the average of the coordinates:

It lies exactly halfway between the two points.

The gradient

The gradient of the segment is the change in $y$ over the change in $x$:

Unlike distance, the order matters here: subtract the coordinates in the same order on top and bottom.

Applying them together

Used together, these formulas verify geometric facts: equal distances show equal sides, equal midpoints show diagonals that bisect each other, and gradient relationships show parallel or perpendicular sides. This is how coordinate geometry proves shapes are squares, parallelograms or right-angled.

Classifying a triangle from its coordinates

Combining the three formulas lets you classify a triangle given its vertices. Compute the three side lengths with the distance formula: all three equal means equilateral, exactly two equal means isosceles, and all different means scalene. To test for a right angle, check whether the gradients of two sides multiply to $-1$, or equivalently whether the side lengths satisfy Pythagoras. So a triangle with sides $5$, $5$ and $\sqrt{50}$ is isosceles, and because $5^2 + 5^2 = 50$, it is also right-angled. Using distance for the sides and gradient (or Pythagoras) for the angles turns a coordinate triangle into a fully classified shape.

Working backwards from a midpoint

A common twist gives you one endpoint and the midpoint, and asks for the other endpoint. Because the midpoint is the average of the endpoints, you rearrange: if $M = (m_x, m_y)$ is the midpoint of $A(x_1, y_1)$ and $B$, then $B = (2m_x - x_1,\ 2m_y - y_1)$. So if $A = (1, 2)$ and the midpoint is $(4, 5)$, then $B = (2(4) - 1, 2(5) - 2) = (7, 8)$. Doubling the midpoint and subtracting the known endpoint reverses the averaging, a neat application of the midpoint formula that appears often in E-Maths problems.

Examples in context

Example 1. Proving a parallelogram. Showing that the diagonals of a quadrilateral share the same midpoint proves it is a parallelogram, because the diagonals of a parallelogram bisect each other. The midpoint formula does the work.

Example 2. The centre of a circle. Given the endpoints of a diameter, the midpoint formula finds the centre of the circle, and the distance formula gives the radius. Two coordinate formulas locate the whole circle.

Try this

Q1. Find the distance between $(0, 0)$ and $(6, 8)$. [2 marks]

• Cue. $\sqrt{6^2 + 8^2} = \sqrt{100} = 10$.

Q2. Find the midpoint of the segment joining $(2, 5)$ and $(8, 1)$. [1 mark]

• Cue. $\left(\dfrac{2 + 8}{2}, \dfrac{5 + 1}{2}\right) = (5, 3)$.

Q3. Find the gradient of the segment joining $(1, 7)$ and $(4, 1)$. [2 marks]

• Cue. $\dfrac{1 - 7}{4 - 1} = \dfrac{-6}{3} = -2$.