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.

Draw a diagramRearrange the given equation into the form y = mx + cy = -3/2x+7/2Gradients of perpendicular lines - m1*m2=-1, so gradient of PQ = 2/3PQ : y=mx+c -> 4=(2/3)*3+cc=2, therefore PQ: y=2/3x+2 b=2/3a+2 since (a,b) lies on PQ