Year 10 Maths Core Straight Line Graphs

Parallel and Perpendicular Lines

Q: Why do perpendicular gradients multiply to -1?

Turning a line through 90 degrees swaps its rise and run and changes one sign. So the new gradient is the negative reciprocal, and their product is -1.

Q: How do you find the gradient of a line in general form?

Make y the subject. For ax + by + c = 0 the gradient is negative a over b.

20 practice questions 1 video lesson Theory + worked examples

Parallel and Perpendicular Lines in NSW Year 10 Core maths links the gradients of two lines with how they meet (NESA outcome MA5‑LIN‑C‑02, with the general-form work from MA5‑LIN‑P‑01). Parallel lines have equal gradients, and perpendicular lines have gradients that multiply to \(-1\), so a line with gradient \(5\) is perpendicular to one with gradient \(-\frac{1}{5}\).

You will find parallel and perpendicular gradients, test whether two lines are perpendicular, find an unknown coefficient so that lines are parallel or perpendicular, and write equations for lines that run parallel or at right angles to a given line. You will also use gradients to show that a quadrilateral is a parallelogram or that a triangle has a right angle, and plan a park footpath that meets another at right angles. It is part of Year 10 Core Straight Line Graphs.

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Theory

This Year 10 maths page covers parallel lines (equal gradients) and perpendicular lines (gradients that multiply to \(-1\)). It shows how to read a gradient from any form of equation, find a missing coefficient, write the equation of a parallel or perpendicular line, and prove a right angle or a parallelogram. The parts in general form matter most if you are heading for Advanced maths in Year 11.

Parallel lines point in the same direction and never meet. They have equal gradients: \(m_1 = m_2\).

Perpendicular lines meet at a right angle (\(90^\circ\)). Their gradients multiply to \(-1\): \(m_1 \times m_2 = -1\). The symbol \(\perp\) means "is perpendicular to".

The perpendicular gradient is the negative reciprocal: flip the fraction and change the sign. The negative reciprocal of \(\dfrac{2}{5}\) is \(-\dfrac{5}{2}\), and of \(-4\) is \(\dfrac{1}{4}\).

In gradient–intercept form \(y = mx + c\), the gradient is \(m\). (Some books write \(y = mx + b\); \(b\) and \(c\) both mean the \(y\)-intercept.) In general form \(ax + by + c = 0\), make \(y\) the subject first to read the gradient.

Horizontal and vertical lines are the special case: a horizontal line (\(m = 0\)) is perpendicular to a vertical line (undefined gradient).

Parallel lines never meet. They have the same gradient: \(m_1 = m_2\).

Perpendicular lines meet at \(90^\circ\). Their gradients multiply to \(-1\): \(2 \times \left(-\dfrac{1}{2}\right) = -1\).

The two tests, for lines with gradients \(m_1\) and \(m_2\):

\[\text{parallel: } m_1 = m_2 \qquad \text{perpendicular: } m_1 \times m_2 = -1, \ \text{so } m_2 = -\dfrac{1}{m_1}\]

m_{1} \times m_{2} = - 1

The point–gradient form gives the line through \((x_1, y_1)\) with gradient \(m\):

\[y - y_1 = m(x - x_1)\]

y - y_{1} = m (x - x_{1})

Gradient from general form. For \(ax + by + c = 0\), \(m = -\dfrac{a}{b}\). For example \(5x + 2y - 1 = 0\) has \(m = -\dfrac{5}{2}\). Rearranging gives the same answer.

An unknown coefficient. Write each gradient in terms of the unknown, then use \(m_1 = m_2\) or \(m_1 m_2 = -1\). For \(2x + ky - 7 = 0\) parallel to \(y = 4x\): \(-\dfrac{2}{k} = 4\), so \(k = -\dfrac{1}{2}\).

Parallelograms. A quadrilateral is a parallelogram if both pairs of opposite sides are parallel: for \(ABCD\), \(m_{AB} = m_{DC}\) and \(m_{AD} = m_{BC}\).

Surd gradients. The negative reciprocal of \(-(\sqrt{2} + 1)\) is \(\dfrac{1}{\sqrt{2} + 1}\). Rationalise: \(\dfrac{1}{\sqrt{2} + 1} \times \dfrac{\sqrt{2} - 1}{\sqrt{2} - 1} = \sqrt{2} - 1\).

How to find a parallel or perpendicular gradient

Find the gradient of the given line: from two points with the gradient formula, or by making \(y\) the subject.
Parallel: use the same gradient.
Perpendicular: flip the fraction and change the sign. Change a decimal to a fraction first.
Check a perpendicular pair: the product must be \(-1\).

How to find the equation of the new line

Find the new gradient \(m\).
Substitute \(m\) and the point into \(y - y_1 = m(x - x_1)\), or into \(y = mx + c\) to find \(c\).
Rearrange to the form the question asks for. For general form, clear fractions and move every term to one side.
Check that the point satisfies your equation.

Example 1 — Parallel and perpendicular gradients

Write the gradient of a line that is parallel and of a line that is perpendicular to a line with gradient:
(i) \(-\dfrac{5}{3}\)
(ii) \(0.4\)

Solution

Parallel: keep the gradient. Perpendicular: flip the fraction and change the sign.

(i)

\(\text{parallel}\)	\(=\)	\(\boldsymbol{-\dfrac{5}{3}}\)
\(\text{perpendicular}\)	\(=\)	\(\boldsymbol{\dfrac{3}{5}}\)

Check: \(-\dfrac{5}{3} \times \dfrac{3}{5} = -1\) ✓

(ii) Write \(0.4\) as \(\dfrac{2}{5}\).

\(\text{parallel}\)	\(=\)	\(\boldsymbol{0.4}\)
\(\text{perpendicular}\)	\(=\)	\(-\dfrac{5}{2} = \boldsymbol{-2.5}\)

m_{⊥} = - 2.5

Example 2 — Lines in general form

(i) Show that \(3x - 4y + 8 = 0\) and \(8x + 6y - 5 = 0\) are perpendicular.
(ii) The line \(y = kx - 2\) is also perpendicular to \(3x - 4y + 8 = 0\). Find \(k\).

Solution

(i) Make \(y\) the subject of each equation to read its gradient.

\(4y\)	\(=\)	\(3x + 8\)
\(y\)	\(=\)	\(\dfrac{3}{4}x + 2 \quad \Rightarrow \quad m_1 = \dfrac{3}{4}\)
\(6y\)	\(=\)	\(-8x + 5\)
\(y\)	\(=\)	\(-\dfrac{4}{3}x + \dfrac{5}{6} \quad \Rightarrow \quad m_2 = -\dfrac{4}{3}\)

\(m_1 \times m_2\)

\(=\)

\(\dfrac{3}{4} \times \left(-\dfrac{4}{3}\right) = -1\)

The product is \(-1\), so the lines are perpendicular.

(ii) The gradient of \(y = kx - 2\) is \(k\).

\(k \times \dfrac{3}{4}\)	\(=\)	\(-1\)
\(k\)	\(=\)	\(\boldsymbol{-\dfrac{4}{3}}\)

k = - \frac{4}{3}

Example 3 — Equation of a perpendicular line

Find the equation of the line through \((-3, 1)\) that is perpendicular to \(3x + 2y - 6 = 0\). Give your answer in general form.

Solution

Find the gradient of the given line, then the perpendicular gradient.

\(2y\)	\(=\)	\(-3x + 6\)
\(y\)	\(=\)	\(-\dfrac{3}{2}x + 3 \quad \Rightarrow \quad m = -\dfrac{3}{2}\)
\(m_\perp\)	\(=\)	\(\dfrac{2}{3}\)

Use the point–gradient form \(y - y_1 = m(x - x_1)\) with \((-3, 1)\):

\(y - 1\)	\(=\)	\(\dfrac{2}{3}(x + 3)\)
\(3y - 3\)	\(=\)	\(2x + 6\)
\(0\)	\(=\)	\(\boldsymbol{2x - 3y + 9}\)

2 x - 3 y + 9 = 0

Check: \(2(-3) - 3(1) + 9 = 0\) ✓. In gradient–intercept form it is \(y = \dfrac{2}{3}x + 3\).

Example 4 — Proving a right angle

Triangle \(PQR\) has vertices \(P(-1, -2)\), \(Q(3, 0)\) and \(R(1, 4)\). Show that the angle at \(Q\) is a right angle.

Solution

The angle at \(Q\) is between the sides \(PQ\) and \(QR\).

\(m_{PQ}\)	\(=\)	\(\dfrac{0 - (-2)}{3 - (-1)} = \dfrac{2}{4} = \dfrac{1}{2}\)
\(m_{QR}\)	\(=\)	\(\dfrac{4 - 0}{1 - 3} = \dfrac{4}{-2} = -2\)
\(m_{PQ} \times m_{QR}\)	\(=\)	\(\dfrac{1}{2} \times (-2) = \boldsymbol{-1}\)

m_{P Q} \times m_{Q R} = - 1

The product is \(-1\), so \(PQ \perp QR\) and the angle at \(Q\) is \(90^\circ\).

Common pitfalls

Do both steps for a perpendicular gradient. Flip and change the sign. The perpendicular to \(m = 3\) is \(-\dfrac{1}{3}\), not \(\dfrac{1}{3}\) or \(-3\).

Make y the subject first. In \(6x - 2y + 5 = 0\) the gradient is not \(6\) or \(-2\). Rearranging gives \(y = 3x + \dfrac{5}{2}\), so \(m = 3\).

Parallel lines are different lines. \(y = 3x + 1\) and \(y = 3x - 4\) are parallel. Equal gradients and equal intercepts would be the same line.

Use the right sides for the angle. To test the angle at \(Q\), use the sides through \(Q\): \(PQ\) and \(QR\).

Frequently asked questions

How do you know if two lines are parallel?

Find the gradient of each line. If the gradients are equal and the lines are different, the lines are parallel.

How do you find a perpendicular gradient?

Take the negative reciprocal: flip the fraction and change its sign. For \(m = \dfrac{2}{3}\), the perpendicular gradient is \(-\dfrac{3}{2}\).

Why do perpendicular gradients multiply to -1?

Turning a line through \(90^\circ\) swaps its rise and run and changes one sign. So the new gradient is the negative reciprocal, and their product is \(-1\).

How do you find the gradient of a line in general form?

Make \(y\) the subject. For \(ax + by + c = 0\) the gradient is \(-\dfrac{a}{b}\).

How do you prove a triangle has a right angle using gradients?

Find the gradients of the two sides that meet at the angle. If they multiply to \(-1\), the sides are perpendicular, so the angle is \(90^\circ\).

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Equation of Lines

Parallel and Perpendicular Lines

📖 Theory

How to find a parallel or perpendicular gradient

How to find the equation of the new line

Common pitfalls

Frequently asked questions

Theory