The point P has coordinates (3,4). The Point Q has the 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.

As we appreciate the rule 1: gradient of line PQ multiplied by the gradient of the line perpendicular to it equals -1. Formula: m x mn=-1We use this to find the gradient of line PQ:Firstly, we find the gradient of the line perpendicular to PQ, mn, by rearranging its equation3x+2y=7
2y=3x-7
y=(3/2)x-7/2
Thus the gradient of the perpendicular line, mn=3/2. Using rule 1 it follows that the gradient of line PQ, m, is equal to -1/mn, which equates to -1/(3/2)= -2/3.Then as we appreciate the rule 2: that the gradient between two points P and Q can be defined as (y2-y1)/(x2-x1), and substitute in the coordinates P (3,4) and Q (a,b). We have the expression (b-4)/(a-3).And using our gradient of PQ, m=-2/3, derived from rule 1, we can equate this expression (b-3)/(a-3) to -2/3, and simply have to rearrange to find b in terms of a.(b-4)/(a-3)=-2/3
b-4=(-2/3)(a-3)
b=(-2/3)(a-3)+4 <---------- Which is our final answer!