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.

The first step should be to rearrange the equation of the perpendicular line to PQ, in order to find the gradient, which works out to -3/2x. This can be done using the equation y=mx+c, where m is the gradient. Then we can use another rule, the fact that if the gradients of two lines are multiplied together, and the answer is -1, that shows that the two lines are in fact perpendicular. Using that concept we can work out the gradient of the line PQ, by calculating -3/2 x m = -1. Here the x represents a multiplication symbol, and m represents the gradient of the line PQ. This works out to be 2/3. The next step is to put the values of P, a coordinate we already know which lies on the line PQ, into the equation of a straight line, y=mx+c, with the gradient we previously worked out, in order to find the "c" in that equation, the y-axis intercept. Hence we will be working out 4=(3*2/3) + c. From this we can work out that c=2. Therefore we have the full equation, y=2/3x + 2. The final answer mark to gain would be to re-write the equation, in terms of b with respect to a, which would be, b=2/3a +2.