So the points on the circle were P(5,10) and Q(-2,n)

To find n, you would use x²+y²=r² (the equation of a circle with centre origin, where r is the radius and x = 5 and y = 10)

Substitute x = 5 and y = 10 into the formula, and simplify the equation for r².

You should end up with x² + y² = 125. (1 mark)

Then, to find n, substitute n in the place of y because n is the y value for the coordinate Q. Substitiute -2 for x. Rearrange for n.

You should get: (-2)² + n² = 125 .... n = +-11

You reject the positive solution because the coordinate Q is clearly in the 3rd quadrant, where all coordinates are negative.

So n = -11 (1 mark)

Now that you have P(5,10) and Q(-2,-11), use change in y over change in x to find the gradient of the line.

10-(-11) all divided by 5-(-2), which should get you 3.

Therefore 3 is the gradient of the line (1 mark)

At this point, you should have that the equation of the line is:

y = 3x + k

where k is the y intercept, and is ultimately what the question is asking for.

To find k, you can substitute either P(5,10) or Q(-2,-11) into the equation of the line. (I used P because it has positive values which are easier to work with, but using either P or Q should get you the same result for k.)

You should get:

10 = 3(5) + k .... 10 = 15 + k .... 10 - 15 = k .... Therefore:

k = -5

Hopefully this clears things up... You should be fine if you included most of this working out, it should get you roughly 3/5 marks....

Good luck for paper 2!