The point P has coordinates (4, 5). The point Q has coordinates (a, b). A line perpendicular to PQ is given by the equation 5x+3y=11. Find an expression for b in terms of a.

First we should find the gradient of the line perpendicular to PQ by rearranging the equation.5x+3y=113y=11-5xy=11/5 -5/3xy= -5/3x + 11gradient is -5/3.We know that the gradients of perpendicular lines multiply together to equal -1. Or, if you prefer, to find the perpendicular gradient, you can 'flip' the first gradient and multiply it by -1. So a gradient of 7/8 is 'flipped' to 8/7 and multiplied by -1 to become -8/7.This means the gradient of PQ is:-5/3 flipped = -3/5, then multiplied by -1 = 3/5.We know that gradient is change in y/change in x. So we need to make an equation we can solve using our new gradient, using the points P and Q:change in y/change in x = 3/55-b/4-a = 3/55(5-b) = 3(4-a)25-5b = 12-3a25+3a = 12+5b13+3a=5bb=3/5 a +13/5