a) A line passes through (0,9) and (3,12) write down the equation of this line . b) A line perpendicular to the line in part a passes through the point (3,14) write the equation of this line.)

a) The gradient of a line is given by the equation (change in y)/(change in x). Therefore to find the gradient of the line in part a we must do (12-9)/(3-0) = 1. Now we have the gradient we can use the formula (y-y1)=m(x-x1). We know m=1 because m stands for the gradient. So we can substitute that into the equation so we have (y-y1)=1(x-x1). Then substitute either of the points into that formula. I will use (3,12) for this example. y-12=1(x-3). When we re-arrange this we can write it as y=x+9.b) We know that the gradient of a line perpendicular to another line is the negative reciprocal of the original line. Therefore we know that the gradient of the line in part b is -1. Similarly to part a we can use the formula (y-y1)=m(x-x1) so we have (y-y1)=-1(x-x1). Then we can substitute the point in part b so we have (y-14)=-1(x-3) which can be rearranged to give y=-x+17.