To write down the equation of a straight line you need two things: a gradient and one point on the line. Every method below is a way of collecting those two ingredients. Writing the equation of a line GivenMethodExample Gradient and one point\(y-y_1=m(x-x_1)\)\(m=2\) through \((3,-1)\) \(y+1=2(x-3)\) \(y=2x-7\) Two pointsFind \(m\) first, then use the point-gradient form with either point.\((1,2)\) and \((4,11)\) \(m=\dfrac{11-2}{4-1}=3\) \(y-2=3(x-1)\) \(y=3x-1\) Gradient and \(y\)-intercept\(y=mx+c\)\(m=-\dfrac{1}{2},\ c=4\) \(y=-\dfrac{1}{2}x+4\) Horizontal and vertical lines through \((h,\,k)\) LineEquationSketch Horizontal line (parallel to the \(x\)-axis)\(y=k\) Vertical line (parallel to the \(y\)-axis)\(x=h\) Perpendicular bisector of \(AB\) Method Find the midpoint of \(AB\) — the bisector passes through it. Find the gradient of \(AB\), then take the perpendicular gradient \(-\dfrac{1}{m_{AB}}\). Substitute the midpoint and the perpendicular gradient into \(y-y_1=m(x-x_1)\). A line can be perpendicular to \(AB\) without cutting it in half, and a line can bisect \(AB\) without meeting it at right angles. The perpendicular bisector does both: it passes through the midpoint \(M\) and is perpendicular to \(AB\). Exam check For a perpendicular bisector, substitute the midpoint — not \(A\) or \(B\) — into the point-gradient form, and use the perpendicular gradient, not \(m_{AB}\).

Equation of a Straight Line and the Perpendicular Bisector

Secondary 3

TGM Original Questions

To write down the equation of a straight line you need two things: a gradient and one point on the line. Every method below is a way of collecting those two ingredients.

Writing the equation of a line

Given	Method	Example
Gradient and one point	\(y-y_1=m(x-x_1)\)	\(m=2\) through \((3,-1)\) \(y+1=2(x-3)\) \(y=2x-7\)
Two points	Find \(m\) first, then use the point-gradient form with either point.	\((1,2)\) and \((4,11)\) \(m=\dfrac{11-2}{4-1}=3\) \(y-2=3(x-1)\) \(y=3x-1\)
Gradient and \(y\)-intercept	\(y=mx+c\)	\(m=-\dfrac{1}{2},\ c=4\) \(y=-\dfrac{1}{2}x+4\)

Horizontal and vertical lines through \((h,\,k)\)

Line	Equation	Sketch
Horizontal line (parallel to the \(x\)-axis)	\(y=k\)
Vertical line (parallel to the \(y\)-axis)	\(x=h\)

Perpendicular bisector of \(AB\)

Method

Find the midpoint of \(AB\) — the bisector passes through it.
Find the gradient of \(AB\), then take the perpendicular gradient \(-\dfrac{1}{m_{AB}}\).
Substitute the midpoint and the perpendicular gradient into \(y-y_1=m(x-x_1)\).

A line can be perpendicular to \(AB\) without cutting it in half, and a line can bisect \(AB\) without meeting it at right angles. The perpendicular bisector does both: it passes through the midpoint \(M\) and is perpendicular to \(AB\).

Exam check

For a perpendicular bisector, substitute the midpoint — not \(A\) or \(B\) — into the point-gradient form, and use the perpendicular gradient, not \(m_{AB}\).

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