Find an equation for the straight line AB , giving your answer in the form px+qy=r, where p, q and r are integers. Given that A has co-ordinates (-2,4) and B has co-ordinates (8,-6)

We know 2 equations for a straight line; Y=MX+C and Y-B=M(X-A). Since we do not have the y intercept (C) or any means of finding it but we do have 2 points on the straight line, it would make sense to use the equation Y -B=M(X-A). Firstly, we need to find the gradient (M), to do this we use the equation M= y1-y2/x1-x2. Let A have x1 (-2)and y1 (4) and B have x2(8) and y2(-6). By subbing these into the gradient equation we get M= 4- (-6) / (-2) -8 which gives us M= 10 / -10, this simplifies to M=-1 ( if the student doesn't understand how the gradient is reached from this equation I would explain it to them here). Next we need to sub the gradient along with 1 of our co-ordinates, lets take A where -2 is A and 4 is B, and then substitute both into our line equation. This gives us : Y- 4 = -1(X- (-2)), the X-(-2) would turn into: X+2, then if we multiply the out the brackets on right hand side we get Y-4 = -X-2. We want the equation in the form pX+ qY=r so we need to put X and Y on one side and the constant on the other. If we move -X to the left and -4 to the right we get: X+Y=-2+4. To finish we just add the numbers to get: X+Y=2, our answer.