**Unformatted text preview: **PERPENDICULAR LINES If two lines of gradient m 1 and 1712 are perpendicular, then the product oftheir
respective gradients must equal —1. ALTERNATIVELY the gradient of one line must equal the negative reciprocal
of the other gradient. The concept is important later, particularly when determining equations of
normals to curves.
7712
These two lines are at
right angles to each other, In X771 2—1.
m1 1 2 Example: Find the gradient of the line which is perpendicular to the line 3/ : inc—5.
3 4
Solution: Z x 2712 : —1 7712 —§ that is, the negative reciprocal. Example: Find the equation of the line passing through (5,—3) and which is
perpendicular to x + 2 y + 6 = 0. Solution: x + 2 y + 6 = 0 {Rearrange into y = mx + b 2y —x—6 —1
y ?x~3 — 1
I711 ""— 2 to find the gradient m 1. Using y — y 1 = m (x — x 1) Since the lines are perpendicular, 2 (x _5) the gradient of this line must be 2, 1
y+3=2x—10 i.e. —§xm2=—1 y = 2x—13 [712 = 2 y-—3 = ...

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