IMAT Mathematics · Geometry

Distance, Midpoint, Gradient and Lines

On a coordinate grid, Pythagoras gives distances, averages give midpoints, and the gradient fixes a line and tells you when two lines are parallel or perpendicular.

All Geometry notes

Why this matters

Lines were a favourite of the older papers: a gradient from two points (2017), a line through a point perpendicular to a given line (2014, 2018), a perpendicular bisector (2011) and the area enclosed by three lines (2022). The ministry papers have asked about circles instead, but every circle question uses distance and midpoint.

Concept 1 of 3: Distance and midpoint between two points

The horizontal and vertical gaps between two points are the legs of a right triangle, and the distance between the points is its hypotenuse. The midpoint sits halfway along both gaps, so its coordinates are the averages of the coordinates of the ends.

Definition

For points $P(x_1, y_1)$ and $Q(x_2, y_2)$ :

Distance $PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .
Midpoint $M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$ .
If $M$ is the midpoint and $P$ one end, the other end is $Q = (2x_M - x_1,\ 2y_M - y_1)$ .

Distance and midpoint

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \qquad M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

Worked example

Find the distance between

A(-2, 5)

and

B(4, -3)

, and the midpoint of AB.

Practice this conceptself-check · 4 quick reps

The midpoint uses sums, the distance uses differences

Average the coordinates for a midpoint:

(x_1 + x_2)/2

. Half the difference,

(x_2 - x_1)/2

, is how far the midpoint is from each end, not where it is.

Concept 2 of 3: Gradient and the equation of a straight line

The gradient says how many units a line rises for each unit it moves to the right. One point and the gradient fix a line completely, so the equation follows straight from them.

Definition

Gradient through $(x_1, y_1)$ and $(x_2, y_2)$ : $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ (rise over run). Positive $m$ slopes up, negative slopes down.
Gradient-intercept form: $y = mx + c$ , with y-intercept $c$ .
Point-gradient form: $y - y_1 = m(x - x_1)$ .
General form $ax + by + c = 0$ : rearrange for $y$ ; the gradient is $-a/b$ .
Horizontal lines are $y = k$ (gradient 0); vertical lines are $x = k$ (no gradient).
Intercepts: put $x = 0$ for the y-intercept and $y = 0$ for the x-intercept.

Gradient and line through a point

m = \frac{y_2 - y_1}{x_2 - x_1} \qquad y - y_1 = m(x - x_1)

Worked example

Find the equation of the line through

(1, -2)

and

(4, 7)

, and its intercepts with the axes.

Practice this conceptself-check · 4 quick reps

Gradient is change in y over change in x

Rise over run:

(y_2 - y_1)/(x_2 - x_1)

. Putting the x-difference on top gives the reciprocal, which IMAT includes as an option. Subtract in the same order on top and bottom.

Concept 3 of 3: Parallel and perpendicular lines, and where lines meet

Parallel lines climb at the same rate, so they have equal gradients. Turning a line through a right angle swaps its rise and run and flips one sign, so perpendicular gradients are negative reciprocals: they multiply to

-1

. Two lines meet where both equations hold at once.

Definition

Parallel lines: $m_1 = m_2$ .
Perpendicular lines: $m_1 m_2 = -1$ , that is $m_2 = -\dfrac{1}{m_1}$ . (A horizontal and a vertical line are also perpendicular.)
The perpendicular bisector of AB passes through the midpoint of AB with the perpendicular gradient. Every point on it is the same distance from A and from B.
The intersection of two lines is found by solving their equations simultaneously.

Parallel and perpendicular gradients

\text{parallel: } m_1 = m_2 \qquad \text{perpendicular: } m_1 m_2 = -1

Worked example

Find the perpendicular bisector of the segment from

A(1, 2)

B(5, 10)

Practice this conceptself-check · 4 quick reps

A perpendicular gradient needs both a flip and a sign change

The perpendicular to a line of gradient

m

has gradient

-1/m

. Taking only the reciprocal (

1/m

) or only the negative (

-m

) gives a line that is not perpendicular; both wrong versions appear among the options.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Distance and midpoint between two points
Distance and midpoint
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \qquad M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
Gradient and the equation of a straight line
Gradient and line through a point
$m = \frac{y_2 - y_1}{x_2 - x_1} \qquad y - y_1 = m(x - x_1)$
Parallel and perpendicular lines, and where lines meet
Parallel and perpendicular gradients
$\text{parallel: } m_1 = m_2 \qquad \text{perpendicular: } m_1 m_2 = -1$

Watch out for (3)

The midpoint uses sums, the distance uses differences
→ Distance and midpoint between two points
Gradient is change in y over change in x
→ Gradient and the equation of a straight line
A perpendicular gradient needs both a flip and a sign change
→ Parallel and perpendicular lines, and where lines meet

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