Point A has coordinates (-1,3). Point B has coordinates (2,-3). Find the equation of the line L that goes through these two points. Point C has coordinates (0,1). Hence or otherwise, find the equation of the line perpendicular to L that goes through C.

First we use the equation gradient = (y1-y2)/(x1-x2) to find the gradient of the line L, using point A (-1,3) and point B (2,-3). This gives us gradient = (3-(-3)/(-1-2) = (3+3)/(-3) = -2. Now we substitute point A into the equation y = mx +c where m is the gradient we just calculated. So 3 = -2*-1+c, so c = 1, and the line equation is y = -2x+1 To find a perpendicular line we use the 'negative reciprocal' of the gradient which means -1/m to find the gradient of the second line. Now we have another gradient, and another point C so we can find the equation of the new line. The new gradient = -1/(-2) = 1/2. So y = 1/2x + c. Using point C (0,1) : 1 = 1/2*0 + c, so c = 1. The second line has equation y = 1/2x + 1