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Year 11 Methods (Unit 1 & 2) Coordinate Geometry And Linear Relations

Parallel And Perpendicular Lines

20 practice questions 1 video lesson Theory + worked examples

Working with parallel and perpendicular lines is a foundational skill assumed for Queensland Year 11 Mathematical Methods (QCAA). Two lines are parallel when their gradients are equal, and perpendicular when their gradients multiply to give negative one.

You will state a parallel or perpendicular gradient using the negative reciprocal, find the equation of such a line through a given point, classify two lines, and solve for an unknown parameter.

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Theory

In Year 11 Mathematical Methods (QCAA), two straight lines are parallel when their gradients are equal, \(m_1=m_2\), and perpendicular when their gradients multiply to \(-1\), \(m_1 m_2=-1\). This page shows how to state a parallel or perpendicular gradient (the negative reciprocal), find the equation of such a line through a point, classify two lines, and solve for an unknown parameter.

Two lines are parallel if they have the same gradient, \(m_1=m_2\). They rise at the same rate and never meet.

Two lines are perpendicular if they meet at a right angle. Their gradients satisfy \(m_1 m_2=-1\), so each is the negative reciprocal of the other: \(m_2=-\dfrac{1}{m_1}\). For example, a gradient of \(\tfrac{1}{2}\) has perpendicular gradient \(-2\).

Parallel: copy the gradient. Perpendicular: flip and negate. To get a perpendicular gradient, turn the fraction upside down and change its sign.
Parallel lines have equal gradientsTwo lines with the same gradient 2 never meet; they are parallel. x y m=2 m=2
Parallel lines share a gradient (\(m_1=m_2\)) and never meet.
Perpendicular linesA line of gradient one half and a line of gradient negative 2 meet at right angles; the product of gradients is negative one. x y m=1/2 m=-2
Perpendicular lines meet at a right angle; \(m_1 m_2=-1\).

Parallel and perpendicular gradient conditions:

\[\text{parallel:}\quad m_1=m_2\]
m1=m2
\[\text{perpendicular:}\quad m_1 m_2=-1\quad\Rightarrow\quad m_2=-\dfrac{1}{m_1}\]
m1m2=-1m2=-1m1
Then use a point. Once you have the gradient, substitute it and the given point into \(y-y_1=m(x-x_1)\) to get the equation.

How to find a parallel or perpendicular line

  1. Gradient of the given line: read \(m\) from \(y=mx+c\) (rearrange first if needed).
  2. New gradient: for parallel keep \(m\); for perpendicular take the negative reciprocal \(-\dfrac{1}{m}\).
  3. Line: substitute the new gradient and the given point into \(y-y_1=m(x-x_1)\) and simplify.
Example 1 — Parallel and perpendicular gradients
A line has equation \(y=3x-7\). State the gradient of a line (a) parallel to it and (b) perpendicular to it.
Solution

Read the gradient of the given line:

\(m_1\)\(=\)\(3\)

(a) Parallel — equal gradients:

\(m_{\parallel}\)\(=\)\(m_1\)
\(=\)\(3\)

(b) Perpendicular — negative reciprocal:

\(m_{\perp}\)\(=\)\(-\dfrac{1}{m_1}\)
\(=\)\(-\dfrac{1}{3}\)

(a) parallel gradient \(3\); (b) perpendicular gradient \(-\dfrac{1}{3}\).

Parallel and perpendicular to y=3x-7A line of gradient 3 with a parallel of gradient 3 and a perpendicular of gradient negative one third. x y m=3 -1/3
m=-13
Example 2 — Equation of a parallel line
Find the equation of the line parallel to \(y=-x+4\) passing through \((1,\,1)\).
Solution

Parallel means the same gradient:

\(m\)\(=\)\(-1\)

Substitute \(m=-1\) and \((1,1)\) into \(y-y_1=m(x-x_1)\):

\(y-1\)\(=\)\(-1(x-1)\)
\(y-1\)\(=\)\(-x+1\)
\(y\)\(=\)\(-x+1+1\)
\(y\)\(=\)\(-x+2\)

Equation: \(y=-x+2\).

Line parallel to y=-x+4 through (1,1)The parallel line y equals negative x plus 2 through the point 1, 1. x y (1,1) y=-x+2
y=-x+2
Example 3 — Equation of a perpendicular line
Find the equation of the line perpendicular to \(y=\dfrac{1}{2}x+1\) passing through \((2,\,2)\).
Solution

Gradient of the given line:

\(m_1\)\(=\)\(\dfrac{1}{2}\)

Perpendicular gradient — negative reciprocal:

\(m\)\(=\)\(-\dfrac{1}{\,1/2\,}\)
\(=\)\(-2\)

Substitute \(m=-2\) and \((2,2)\):

\(y-2\)\(=\)\(-2(x-2)\)
\(y-2\)\(=\)\(-2x+4\)
\(y\)\(=\)\(-2x+6\)

Equation: \(y=-2x+6\).

Line perpendicular to y=x/2+1 through (2,2)The perpendicular line y equals negative 2x plus 6 crossing the line of gradient one half at 2, 2. x y (2,2) y=-2x+6
y=-2x+6
Example 4 — Finding an unknown gradient parameter
For what value of \(k\) is the line \(y=kx+1\) perpendicular to \(y=\dfrac{1}{4}x-2\)?
Solution

Perpendicular gradients multiply to \(-1\):

\(k\times\dfrac{1}{4}\)\(=\)\(-1\)

Multiply both sides by \(4\):

\(k\)\(=\)\(-1\times 4\)
\(k\)\(=\)\(-4\)

Parameter: \(k=-4\).

Finding k for perpendicularityA line of gradient one quarter and a perpendicular line of gradient negative 4 through 0, 1. x y m=1/4 k=-4
k=-4

Common pitfalls

Forgetting the negative in a perpendicular gradient. It is the negative reciprocal \(-\dfrac{1}{m}\) — not just \(\dfrac{1}{m}\), and not \(-m\).
Reading the gradient before rearranging. If a line is given as \(x+2y=6\), rearrange to \(y=-\tfrac{1}{2}x+3\) so the gradient is \(-\tfrac{1}{2}\) before comparing.
Using the wrong point. The new line passes through the given point, not a point on the original line. Substitute the stated point into the point-gradient form.

Frequently asked questions

When are two lines parallel?

When their gradients are equal, \(m_1=m_2\). Parallel lines rise at the same rate and never intersect.

When are two lines perpendicular?

When their gradients multiply to \(-1\), \(m_1 m_2=-1\); each gradient is the negative reciprocal of the other.

What is the perpendicular gradient of a line with gradient 2/3?

The negative reciprocal: flip \(\tfrac{2}{3}\) to \(\tfrac{3}{2}\) and negate, giving \(-\tfrac{3}{2}\).

How do you find the equation of a perpendicular line through a point?

Take the negative reciprocal gradient, then substitute it and the point into \(y-y_1=m(x-x_1)\) and simplify.

Are the axes perpendicular under this rule?

They are perpendicular, but the rule \(m_1 m_2=-1\) does not apply directly: the \(y\)-axis is vertical with an undefined gradient, a special case.