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Before you start, it may be helpful to read the guides on coordinates and graphs from and .

Reminder

• The line \(y = mx +c\) has a gradient \(m\) and an intercept \(c\)
• Parallel lines have the same gradient

Line A \(m=–\frac{1}{2}\)

Line B \(m=2\)

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Product of gradients \(=–1\)

Example

A line L₁ has the equation \(3y = 2x - 5\).What is the equation of the line L₂ which goes through (4, -4) and is at right angles to L₁?

Solution:

Divide L₁ by 3 to get \(y = \frac{2}{3}x -\frac{5}{3}\)

\(m = \frac{2}{3}\)

For L₂

\(m = - \frac{3}{2}\)

To find the gradient of a perpendicular line, turn the given gradient upside down and put a minus sign in front.

Let equation of L₂ be \(y = mx + c\)
\(y = - \frac{3}{2}x + c\)

Goes through (4, -4) substitute x = 4, y = -4
\(-4 = - \frac{3}{2}(4) + c\)
\(-4 = -6 + c\)
\(c = 2\)

Equation is \(y = - \frac{3}{2}x + 2\)

It is usual to eliminate fractions from the equation of a line by multiplying across by the denominator of the fraction.

\(2y = -3x + 4\)
or \(2y + 3x = 4\)

Alternative solution:

Use the formula \(y - y_1 = m (x - x_1)\) for L₂

\(m = - \frac{3}{2}\)
\(y - (-4) = - \frac{3}{2}(x -4)\)
\(y + 4 = - \frac{3}{2}(x -4)\)

Multiply across by 2

\(2y + 8 = - 3(x -4)\)
\(2y + 8 = - 3x + 12\)

Equation of L₂ \(2y + 3x = 4\)

Answer:

Mid point of the line (5,5) and (-5, -1) is (0, 2)

\(\matrix{~x & ~y \cr ~5 & ~5 \cr -5 & -1 \cr \hline ~0 & ~4}\)

Divide each by 2 to give mid point of (0, 2).

Let equation of the line which is the perpendicular bisector be \(y = mx + c\)
It goes through (0, 2).

Gradient of line joining (5,5) and (-5, -1) = \(\frac{\text{difference in y values}}{\text{difference in x values}} = \frac{5-(-1)}{5-(-5)} = \frac{6}{10}\)

Gradient of perpendicular line = \( - \frac{10}{6} \)

\(m = - \frac{10}{6} \)
It goes through (0, 2).

Substitute x = 0, y = 2 and \(m = - \frac{10}{6} \)
\(2=- \frac{10}{6}(0) + c\)
\(c = 2\)

Equation is \(y =- \frac{10}{6}x + 2\)

Multiply by 6
\(6y =-10x + 12\)