Meeting at a right angle
Two lines are perpendicular when they cross at \(90^\circ\). Unlike parallel lines, which share a gradient, perpendicular lines have gradients that are tied together by a fixed relationship.
- Perpendicular lines
- Two lines that meet at a right angle. If their gradients are \(m_1\) and \(m_2\), then \(m_1 m_2 = -1\), provided neither line is vertical.
The negative reciprocal rule
The gradients of perpendicular lines multiply to \(-1\):
Rearranged, the second gradient is the negative reciprocal of the first:
So gradient \(2\) pairs with \(-\tfrac{1}{2}\), and gradient \(-\tfrac{3}{4}\) pairs with \(+\tfrac{4}{3}\). Check: \(2 \times \left(-\tfrac{1}{2}\right) = -1\).
Negative reciprocal in two steps
Flip the fraction, then swap the sign. Gradient \(5\) becomes \(-\tfrac{1}{5}\); gradient \(-\tfrac{2}{7}\) becomes \(+\tfrac{7}{2}\). Doing only one step is the classic slip in Criteria A and C tasks.
Finding a perpendicular gradient
Given any gradient, you can write down the perpendicular one at once. Whole numbers count as fractions over \(1\), so gradient \(3 = \tfrac{3}{1}\) flips to \(\tfrac{1}{3}\) and becomes \(-\tfrac{1}{3}\).
Worked example
Find the equation of the line perpendicular to \(y = 2x + 1\) that passes through \((4, 3)\).
Check yourself
1. What gradient is perpendicular to \(y = 5x - 2\)? +
Take the negative reciprocal of \(5 = \tfrac{5}{1}\): flip to \(\tfrac{1}{5}\), change sign to get \(\mathbf{-\tfrac{1}{5}}\).
2. Is \(y = 2x + 1\) perpendicular to \(y = -\tfrac{1}{2}x + 4\)? +
Multiply the gradients: \(2 \times \left(-\tfrac{1}{2}\right) = -1\). Because the product is \(-1\), yes, they are perpendicular.
3. Find the line perpendicular to \(y = -3x\) through \((0, 1)\). +
Negative reciprocal of \(-3\) is \(\tfrac{1}{3}\). The point \((0, 1)\) gives \(c = 1\), so the equation is \(\mathbf{y = \tfrac{1}{3}x + 1}\).
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