Perpendicular and Parallel Lines
Every question with a fully worked solution.
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Theory
Parallel lines have equal gradients, and perpendicular lines have gradients that multiply to \(-1\). This Year 10A maths page covers testing lines in any form, finding unknown coefficients (including ones that lead to a quadratic), the equation of a perpendicular line and the shortest distance from a point to a line, and using gradients to prove right angles in shapes.
Two lines are parallel if they never meet. They rise at the same rate, so they have the same gradient: \(m_1 = m_2\). Lines with the same gradient and the same \(y\)-intercept are the same line, not two parallel lines.
Two lines are perpendicular if they meet at a right angle. Their gradients multiply to \(-1\): \(m_1 m_2 = -1\). So each gradient is the negative reciprocal of the other: flip the fraction and change the sign. The negative reciprocal of \(\dfrac{2}{3}\) is \(-\dfrac{3}{2}\).
A horizontal line (gradient \(0\)) is perpendicular to a vertical line (undefined gradient). This is the one case the rule \(m_1 m_2 = -1\) does not cover.
The shortest distance from a point to a line is along the perpendicular. The point where the perpendicular meets the line is called the foot of the perpendicular.
Gradient of a line in general form \(ax + by + c = 0\) (with \(b \ne 0\)):
Equation of a line through \((x_1,\ y_1)\) with gradient \(m\):
How to test or use parallel and perpendicular lines
- Find each gradient: rearrange to \(y = mx + c\), use \(m = -\dfrac{a}{b}\), or use two points or the two intercepts.
- Compare: equal gradients mean parallel; a product of \(-1\) means perpendicular.
- Write the new line with \(y - y_1 = m(x - x_1)\), using the same gradient (parallel) or the negative reciprocal (perpendicular).
Problems that need an equation
- Unknown coefficient: write the gradient in terms of \(k\), then set \(m_1 = m_2\) or \(m_1 m_2 = -1\) and solve (Example 2). If \(k\) appears in both gradients you may get a quadratic; check each answer gives two different lines.
- Unknown point: if a line through \((k,\ 3)\) and another point must be perpendicular to a given line, set its gradient equal to the negative reciprocal and solve.
- Right angle in a triangle: the two sides at the right-angle vertex are perpendicular. If the vertex itself is unknown, such as \((c,\ 0)\), the equation \(m_1 m_2 = -1\) becomes a quadratic with two answers.
- Lines meeting on an axis: meeting on the \(y\)-axis means equal \(y\)-intercepts.
- Foot of the perpendicular and shortest distance: find the perpendicular line, solve it simultaneously with the given line, then use the distance formula (Example 3).
- Shapes: parallel sides give parallelograms and trapeziums; a product of \(-1\) gives a right angle, so a rectangle or a right-angled trapezium (Example 4).
- Areas with intercepts: a line meeting the axes at \(P\) and \(Q\) forms a right-angled triangle \(OPQ\) with area \(\dfrac{1}{2} \times OP \times OQ\). This also gives a check on a perpendicular distance.
Write each line in the form \(y = mx + c\) to read off its gradient.
| \(\ell_1:\ 6y\) | \(=\) | \(4x + 1\) |
| \(y\) | \(=\) | \(\dfrac{2}{3}x + \dfrac{1}{6}\) |
| \(m_1\) | \(=\) | \(\dfrac{2}{3}\) |
| \(\ell_2:\) | \(\) | \(m_2 = \dfrac{2}{3}\) |
| \(\ell_3:\ 2y\) | \(=\) | \(-3x + 8\) |
| \(y\) | \(=\) | \(-\dfrac{3}{2}x + 4\) |
| \(m_3\) | \(=\) | \(-\dfrac{3}{2}\) |
\(m_1 = m_2\) and the \(y\)-intercepts differ, so \(\ell_1 \parallel \ell_2\).
| \(m_1 \times m_3\) | \(=\) | \(\dfrac{2}{3} \times \left(-\dfrac{3}{2}\right) = -1\) |
So \(\ell_1 \parallel \ell_2\), and \(\ell_3\) is perpendicular to both.
(i) parallel
(ii) perpendicular.
Make \(y\) the subject of the first equation to get its gradient in terms of \(k\).
| \(ky\) | \(=\) | \(-3x + 4\) |
| \(y\) | \(=\) | \(-\dfrac{3}{k}x + \dfrac{4}{k},\quad m = -\dfrac{3}{k}\) |
(i) Parallel: the gradients are equal.
| \(-\dfrac{3}{k}\) | \(=\) | \(2\) |
| \(-3\) | \(=\) | \(2k\) |
| \(k\) | \(=\) | \(\mathbf{-\dfrac{3}{2}}\) |
(ii) Perpendicular: the gradients multiply to \(-1\).
| \(-\dfrac{3}{k} \times 2\) | \(=\) | \(-1\) |
| \(-6\) | \(=\) | \(-k\) |
| \(k\) | \(=\) | \(\mathbf{6}\) |
Check (ii): \(k = 6\) gives gradient \(-\dfrac{1}{2}\), and \(-\dfrac{1}{2} \times 2 = -1\ \checkmark\)
(i) Find the equation of the line through \(P\) perpendicular to \(\ell\).
(ii) Find \(N\), where it meets \(\ell\).
(iii) Find the exact shortest distance from \(P\) to \(\ell\).
(i) \(\ell\) is \(y = -\dfrac{1}{2}x + 2\), so its gradient is \(-\dfrac{1}{2}\) and the perpendicular gradient is \(2\). Use the point–gradient form.
| \(y - 6\) | \(=\) | \(2(x - 1)\) |
| \(y\) | \(=\) | \(2x + 4\) |
(ii) Substitute \(y = 2x + 4\) into \(x + 2y - 4 = 0\).
| \(x + 2(2x + 4)\) | \(=\) | \(4\) |
| \(5x + 8\) | \(=\) | \(4\) |
| \(x\) | \(=\) | \(-\dfrac{4}{5}\) |
| \(y\) | \(=\) | \(2\left(-\dfrac{4}{5}\right) + 4\) |
| \(\) | \(=\) | \(\dfrac{12}{5}\) |
So \(\boldsymbol{N\left(-\dfrac{4}{5},\ \dfrac{12}{5}\right)}\).
(iii) The shortest distance is the perpendicular distance \(PN\).
The run is \(1 - \left(-\dfrac{4}{5}\right) = \dfrac{9}{5}\) and the rise is \(6 - \dfrac{12}{5} = \dfrac{18}{5}\).
| \(PN^2\) | \(=\) | \(\left(\dfrac{9}{5}\right)^2 + \left(\dfrac{18}{5}\right)^2\) |
| \(\) | \(=\) | \(\dfrac{81}{25} + \dfrac{324}{25}\) |
| \(\) | \(=\) | \(\dfrac{405}{25}\) |
| \(PN\) | \(=\) | \(\dfrac{\sqrt{405}}{5}\) |
| \(\) | \(=\) | \(\dfrac{9\sqrt{5}}{5}\text{ units}\) |
A rectangle is a parallelogram with a right angle. Find all four gradients.
| \(m_{AB}\) | \(=\) | \(\dfrac{0 - (-2)}{3 - (-1)} = \dfrac{2}{4} = \dfrac{1}{2}\) |
| \(m_{DC}\) | \(=\) | \(\dfrac{2 - 0}{2 - (-2)} = \dfrac{2}{4} = \dfrac{1}{2}\) |
| \(m_{BC}\) | \(=\) | \(\dfrac{2 - 0}{2 - 3} = -2\) |
| \(m_{AD}\) | \(=\) | \(\dfrac{0 - (-2)}{-2 - (-1)} = -2\) |
Both pairs of opposite sides are parallel, so \(ABCD\) is a parallelogram. Now test one angle:
| \(m_{AB} \times m_{BC}\) | \(=\) | \(\dfrac{1}{2} \times (-2) = -1\) |
\(AB \perp BC\), so \(\angle B = 90^\circ\). A parallelogram with a right angle has four right angles: \(ABCD\) is a rectangle.
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 not the same line, they are parallel.
How do you find the gradient of a perpendicular line?
Take the negative reciprocal: flip the gradient upside down and change its sign. A line perpendicular to one with gradient \(2\) has gradient \(-\dfrac{1}{2}\).
Why do perpendicular gradients multiply to negative 1?
Turning a line through \(90^\circ\) swaps its rise and run and reverses one direction, so a gradient of \(\dfrac{\text{rise}}{\text{run}}\) becomes \(-\dfrac{\text{run}}{\text{rise}}\). Their product is \(-1\).
How do you find the equation of a line perpendicular to another line through a point?
Find the gradient of the given line, take its negative reciprocal, then substitute that gradient and the point into \(y - y_1 = m(x - x_1)\).
How do you find the gradient of ax + by + c = 0?
Rearrange to make \(y\) the subject. The gradient is \(-\dfrac{a}{b}\). For \(3x + 4y - 1 = 0\) the gradient is \(-\dfrac{3}{4}\).