Linear Graphs
Coordinate Geometry and Graphs
Negative reciprocal gradients
- Two lines are perpendicular when they cross at a right angle
- Their gradients multiply to give −1, so each is the negative reciprocal of the other
- 4 pairs with −0.25, and ⅔ pairs with −1.5
- To turn one gradient into the other, flip the fraction over and change the sign
- Rearrange both equations to y = mx + c before comparing, otherwise the gradients cannot be read
- The special case is a horizontal and a vertical line: y = c and x = k are perpendicular, and this is the one pair whose gradients cannot be multiplied
Building the equation
- Rearrange the given line and read its gradient
- Take the negative reciprocal to get the gradient of the new line
- Substitute the point you are given to find c, exactly as for a parallel line
Worked example
Perpendicular through a given point
A line passes through (6, 1) and is perpendicular to y = ⅔x + 4. Find its equation. Give your answer in the form y = mx + c.
Solution:
- The given gradient is ⅔
- Flip it and change the sign: the new gradient is −1.5
- Check: ⅔ × (−1.5) = −1
- Start from y = −1.5x + c
- Substitute x = 6 and y = 1: 1 = −9 + c
- Add 9: c = 10
- The equation is y = −1.5x + 10
The perpendicular bisector
- Bisecting a segment at right angles means passing through its midpoint at 90°, and the line that does so is its perpendicular bisector
- It needs both skills together: the midpoint gives the point it passes through, and the negative reciprocal gives its gradient
- Find the midpoint, find the gradient of the segment, flip and negate that gradient, then substitute the midpoint to find c
Exam tip
Write each step of a perpendicular line
Perpendicular lines, including perpendicular bisectors, are set in 8 of the 21 papers. Write the first gradient, then −1 ÷ it, then the point substituted into y = mx + c, each on its own line. In all eight schemes the last two lines earn a method mark each, and the substitution mark follows through from your own perpendicular gradient, so an early slip does not cost it. A gradient worked out from two points earns its own mark too.
Worked example
Perpendicular bisector of a segment
Find an equation for the perpendicular bisector of the segment from P(2, −1) to Q(10, 5).
Solution:
- Midpoint: ((2 + 10) ÷ 2, (−1 + 5) ÷ 2) = (6, 2)
- Gradient of PQ: (5 − (−1)) ÷ (10 − 2) = 6 ÷ 8 = 0.75
- Negative reciprocal: −4 ÷ 3
- Start from y = (−4 ÷ 3)x + c
- Substitute (6, 2): 2 = −8 + c, so c = 10
- The perpendicular bisector is y = (−4 ÷ 3)x + 10