Parallel and perpendicular lines

Parallel lines share a gradient; perpendicular gradients multiply to −1 (the negative reciprocal). Using both — and the perpendicular bisector of a segment — to find equations of lines.

Pure · Coordinate geometry

Parallel and perpendicular lines

Parallel lines share a gradient; perpendicular gradients multiply to −1 (the negative reciprocal). Using both — and the perpendicular bisector of a segment — to find equations of lines.

Why it works

Two lines are parallel when they have the same gradient — same steepness, never meeting. So to build a line parallel to a given one, lift its gradient straight off and pair it with the new point.

Two lines are perpendicular when their gradients multiply to

-1

m_1 m_2 = -1, \qquad \text{equivalently} \qquad m_2 = -\frac{1}{m_1}.

That second form, the negative reciprocal, is the one you use in practice: flip it and change the sign — both steps. The reason is geometric: rotate a gradient triangle by

90^\circ

and its rise and run swap roles while one of them reverses direction, so

\frac{\text{rise}}{\text{run}}

becomes

-\frac{\text{run}}{\text{rise}}

. The two slips that hide here are doing only one step. For

m_1 = 2

the perpendicular gradient is

-\tfrac12

; writing

\tfrac12

(flipped but not negated) or

-2

(negated but not flipped) is wrong. Always check the product comes to

-1

2 \times -\tfrac12 = -1

. ✓

The one exception is a horizontal or vertical line. A horizontal line (

m = 0

) and a vertical line (gradient undefined) are perpendicular, but you can't get there through "

-\frac1m

" because you'd divide by zero — recognise the case and write

x = k

y = k

directly.

A perpendicular bisector of a segment

AB

combines both ideas: it cuts

AB

in half at right angles, so it (1) passes through the midpoint of

AB

and (2) has gradient the negative reciprocal of

AB

's gradient. Every point on it is equidistant from

A

and

B

— which is exactly why it's the key to finding the centre of a circle later.