Year 10 Maths - 10 Linear Relations

Perpendicular and Parallel 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

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

Watch 1 video(s)

Perpendicular and Parallel Lines - Video - Parallel and Perpendicular Lines Watch

Theory

Parallel lines have equal gradients and perpendicular lines have gradients that multiply to \(-1\). This Year 10 maths page shows how to read a gradient from any form of equation, find a missing coefficient or coordinate, write the equation of a parallel or perpendicular line, and use gradients to prove right angles and find the shapes and areas that lines enclose.

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 \(m_2\):

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

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 \(y = 5x\): \(-\dfrac{2}{k} = 5\), so \(k = -\dfrac{2}{5}\).

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

An unknown in both points. The line through \((k,\ 1)\) and \((5,\ k)\) is perpendicular to a line with gradient \(\dfrac{1}{2}\), so its gradient is \(-2\). Write the gradient formula with the unknown and solve: \(\dfrac{k - 1}{5 - k} = -2\), so \(k - 1 = -10 + 2k\) and \(k = 9\). Check: \((9,\ 1)\) and \((5,\ 9)\) give \(\dfrac{9 - 1}{5 - 9} = -2\).

The unknown in both gradients. If \(y = (k + 1)x + 3\) and \(y = (k - 1)x - 2\) are perpendicular, then \((k + 1)(k - 1) = -1\). Expand: \(k^2 - 1 = -1\), so \(k^2 = 0\) and \(k = 0\). Check: the gradients are \(1\) and \(-1\), and \(1 \times (-1) = -1\). An equation like this can be a quadratic, so solve it fully and check every answer.

Lines that meet on an axis. Two lines that meet on the \(y\)-axis have the same \(y\)-intercept; two lines that meet on the \(x\)-axis have the same \(x\)-intercept. Use this as a second equation when a question has two unknowns.

Triangles and trapeziums. To find the corners of a triangle made by three lines, solve the equations two at a time. The lines \(y = x + 1\), \(y = -x + 7\) and \(y = 1\) meet at \((3,\ 4)\), \((0,\ 1)\) and \((6,\ 1)\). The first two have gradients \(1\) and \(-1\), so the right angle is at \((3,\ 4)\). With base \(6\) along \(y = 1\) and height \(4 - 1 = 3\), the area is \(\dfrac{1}{2} \times 6 \times 3 = 9\) square units. A trapezium has one pair of parallel sides; it is right-angled if another side is perpendicular to them.

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 \(2x - 5y + 4 = 0\) and \(10x + 4y - 3 = 0\) are perpendicular.
(ii) The line \(y = kx - 2\) is also perpendicular to \(2x - 5y + 4 = 0\). Find \(k\).

Solution

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

\(5y\)	\(=\)	\(2x + 4\)
\(y\)	\(=\)	\(\dfrac{2}{5}x + \dfrac{4}{5}\)
\(m_1\)	\(=\)	\(\dfrac{2}{5}\)
\(4y\)	\(=\)	\(-10x + 3\)
\(y\)	\(=\)	\(-\dfrac{5}{2}x + \dfrac{3}{4}\)
\(m_2\)	\(=\)	\(-\dfrac{5}{2}\)

\(m_1 \times m_2\)	\(=\)	\(\dfrac{2}{5} \times \left(-\dfrac{5}{2}\right)\)
\(\)	\(=\)	\(-1\)

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

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

\(k \times \dfrac{2}{5}\)	\(=\)	\(-1\)
\(k\)	\(=\)	\(\boldsymbol{-\dfrac{5}{2}}\)

k = - \frac{5}{2}

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.

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

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 \(90^\circ\).

← Previous subtopic

Length and Midpoint of a Line Segment

Next subtopic →

Simultaneous Equations Using Substitution

Perpendicular and Parallel 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