The point P has coordinates (3, 4) The point Q has coordinates (a, b) A line perpendicular to PQ is given by the equation 3x + 2y = 7 Find an expression for b in terms of a
Perpendicular gradients multiply to give -1. The gradient of the perpendicular line (y=-3/2x+7/2) is -3/2 , so the gradient of PQ is 2/3. Using gradient formula change in y/ change in x :(4-b)/(3-a) = 2/3 Simplifying:4-b=(6-2a)/3 12-3b=6-2a 3b=6+2a In terms of b: b=2+2/3a
LD
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