P has coordinates (3,4), Q has coordinate (a,b), a line perpendicular to PQ has equation 3x+2y=7. Find an expression for b in terms of a

Rearrange the equation of the line perpendicular to PQ to give y = -(3/2)x + 7/2. Gradient of this line = -3/2Using the knowledge that the gradients of a line perpendicular to another line is related by a factor of -1/x we know the gradient of PQ is 2/3The gradient is the (difference in y)/(difference in x) which is (b-4)/(a-3) = 2/3multiply the bottom of the LHS to get b-4 = 2a/3 - 2 and take the 4 over to get b = 2a/3 + 2.