Parallel and Perpendicular Lines
Every question with a fully worked solution.
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Theory
This Year 9 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.
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).
The two tests, for lines with gradients \(m_1\) and \(m_2\):
The point–gradient form gives the line through \((x_1, y_1)\) with gradient \(m\):
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.
(i) \(-\dfrac{5}{3}\)
(ii) \(0.4\)
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 \(\dfrac{2}{5}\).
| \(\text{parallel}\) | \(=\) | \(\boldsymbol{0.4}\) |
| \(\text{perpendicular}\) | \(=\) | \(-\dfrac{5}{2} = \boldsymbol{-2.5}\) |
(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 \(k\).
(i) Make \(y\) the subject of each equation to read its gradient.
| \(4y\) | \(=\) | \(3x + 8\) |
| \(y\) | \(=\) | \(\dfrac{3}{4}x + 2\) |
| \(m_1\) | \(=\) | \(\dfrac{3}{4}\) |
| \(6y\) | \(=\) | \(-8x + 5\) |
| \(y\) | \(=\) | \(-\dfrac{4}{3}x + \dfrac{5}{6}\) |
| \(m_2\) | \(=\) | \(-\dfrac{4}{3}\) |
| \(m_1 \times m_2\) | \(=\) | \(\dfrac{3}{4} \times \left(-\dfrac{4}{3}\right) = -1\) |
The product is \(-1\), so the lines are perpendicular.
(ii) The gradient of \(y = kx - 2\) is \(k\).
| \(k \times \dfrac{3}{4}\) | \(=\) | \(-1\) |
| \(k\) | \(=\) | \(\boldsymbol{-\dfrac{4}{3}}\) |
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 \((-3, 1)\):
| \(y - 1\) | \(=\) | \(\dfrac{2}{3}(x + 3)\) |
| \(3y - 3\) | \(=\) | \(2x + 6\) |
| \(0\) | \(=\) | \(\boldsymbol{2x - 3y + 9}\) |
Check: \(2(-3) - 3(1) + 9 = 0\) ✓. In gradient–intercept form it is \(y = \dfrac{2}{3}x + 3\).
The angle at \(Q\) is between the sides \(PQ\) and \(QR\).
| \(m_{PQ}\) | \(=\) | \(\dfrac{0 - (-2)}{3 - (-1)}\) |
| \(\) | \(=\) | \(\dfrac{2}{4} = \dfrac{1}{2}\) |
| \(m_{QR}\) | \(=\) | \(\dfrac{4 - 0}{1 - 3}\) |
| \(\) | \(=\) | \(\dfrac{4}{-2} = -2\) |
| \(m_{PQ} \times m_{QR}\) | \(=\) | \(\dfrac{1}{2} \times (-2)\) |
| \(\) | \(=\) | \(\boldsymbol{-1}\) |
The product is \(-1\), so \(PQ \perp QR\) and the angle at \(Q\) is \(90^\circ\).
Common pitfalls
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 \(90^\circ\).