Find the gradient, length and midpoint of a line segment, write the equation of a line, and use the conditions for parallel and perpendicular lines

What this dot point is asking

SEAB wants you to work confidently with points and lines on the Cartesian plane: to find the gradient, length and midpoint of a segment, to write the equation of a line in various forms, and to use the gradient conditions that detect parallel and perpendicular lines. These tools are the backbone of every coordinate-geometry problem.

The answer

Gradient, distance and midpoint

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

The distance formula is Pythagoras applied to the horizontal and vertical gaps.

The equation of a line

The two most useful forms:

Use point-gradient when you know a point and the gradient; rearrange to gradient-intercept to read off the $y$-intercept.

Parallel and perpendicular lines

Two lines are parallel when their gradients are equal, and perpendicular when the product of their gradients is $-1$:

So the gradient perpendicular to $m$ is the negative reciprocal $-\dfrac{1}{m}$.

The perpendicular bisector

A perpendicular bisector of a segment passes through the midpoint and is perpendicular to the segment. Find the midpoint, take the negative reciprocal of the segment's gradient, and write the line through the midpoint.

Finding the foot of the perpendicular

A common A-Maths extension is to find the foot of the perpendicular from a point to a line, the closest point on the line. The method chains the tools: take the negative reciprocal of the line's gradient to get the perpendicular's gradient, write the perpendicular through the given point, then solve it simultaneously with the original line to find their intersection. That intersection is the foot, and the distance from the point to it is the shortest distance to the line. Setting up the perpendicular and intersecting it with the line is the standard route, and it reuses the gradient, equation, and simultaneous-solving skills together.

Using the ratio of a point dividing a segment

Beyond the midpoint, A-Maths asks for a point dividing a segment in a given ratio. A point $P$ dividing $AB$ in the ratio $m : n$ has coordinates $\left(\tfrac{n x_1 + m x_2}{m + n}, \tfrac{n y_1 + m y_2}{m + n}\right)$, which reduces to the midpoint formula when $m = n$. So the point dividing $A(1, 2)$ and $B(7, 8)$ in the ratio $2 : 1$ is $\left(\tfrac{1(1) + 2(7)}{3}, \tfrac{1(2) + 2(8)}{3}\right) = (5, 6)$. Recognising the midpoint as the special case $1 : 1$ of this general section formula ties the two ideas together and handles the ratio problems that go beyond simple bisection.

Examples in context

Example 1. Shortest distance to a path. The shortest route from a point to a straight road is along the perpendicular from the point to the line; finding that perpendicular and its foot uses the negative-reciprocal gradient, a classic optimisation in surveying.

Example 2. Detecting a right angle in a shape. To confirm a triangle has a right angle at a vertex, check that the gradients of the two sides meeting there multiply to $-1$, turning a geometric claim into a quick gradient calculation.

Try this

Q1. Find the gradient of the line joining $(1, 4)$ and $(3, 10)$. [2 marks]

• Cue. $\dfrac{10 - 4}{3 - 1} = 3$.

Q2. State the gradient of a line perpendicular to $y = 2x + 1$. [1 mark]

• Cue. Negative reciprocal of $2$ is $-\dfrac{1}{2}$.

Q3. Find the equation of the line through $(0, 3)$ parallel to $y = 4x - 7$. [2 marks]

• Cue. Same gradient $4$, intercept $3$: $y = 4x + 3$.