Cambridge Pre-U Mathematics Thus the perpendicular distance d is given by the formula 2 2 b2 u − abv − ac a2 v − abu − bc 2 = u− + v− d a2 + b 2 a2 + b 2 1 2 2 2 2 = u + abv + ac) + (b v + abu + bc) (a (a2 + b2 )2 1 2 2 2 2 (au + bv + c) + b (au + bv + c) = a (a2 + b2 )2 =
(au + bv + c)2 a2 + b 2
To calculate d, we need to take the positive square root of the right-hand side. Key Fact 2.9 Perpendicular Distance to a Line The perpendicular distance from a point (u, v) to the line ax + by + c = 0 is au + bv + c √ a2 + b 2 (or minus the above if this expression is negative).
Looking ahead to Chapter 4, we use the modulus function to write this expression as |au + bv + c| √ a2 + b 2
Exercise 2C
1. In each part write down the gradient of a line which is perpendicular to one with the given gradient.
a) 2 e) −1 p i) q
b) −3 f) 1 34 j) 0
c) 4 1 g) − m k) −m
d) − 56 h) m a l) b−c
2. In each part find the equation of the line through the given point which is perpendicular to the given line. Write your final answer so that it doesn’t contain fractions.
a) c) e) g) i) k)
(2, 3), y = 4x + 3 (2, −5), y = −5x − 2 (−1, 4), 2x + 3y = 8 (5, −3), 2x = 3 (0, 0), y = mx + c (c, d), ny − x = p
b) d) f) h) j) l)
(−3, 1), y = − 12 x + 3 (7, −4), y = 2 12 (4, 3), 3x − 5y = 8 (0, 3), y = 2x − 1 (a, b), y = mx + c (−1, −2), ax + by = c
3. Find the equation of the line through the point (−2, 5) which is perpendicular to the line y = 3x + 1. Find also the point of intersection of the two lines. 4. Find the equation of the line through the point (1, 1) which is perpendicular to the line 2x − 3y = 12. Find also the point of intersection of the two lines. 5. Find the angle between the lines y = 2x − 1 and y = 12 x + 7 to 2 decimal places. 6. A line through a vertex of a triangle which is perpendicular to the opposite side is called an altitude. Find the equation of the altitude through the vertex A of the triangle ABC where A is the point (2, 3), B is (1, −7) and C is (4, −1). 7. P (2, 5), Q (12, 5) and R (8, −7) form a triangle. 28