Exam Questions


January 2005



We want to start by looking at the equation we've been given and seeing that it's in y = mx + c format. Parallel lines have the same gradient (represented by m in the equation), so this is where we can start. We also know that L passes through the point (5, 2), so we know we have enough information to write the equation in y - y1 = m(x - x1).

m = -2, therefore the equation of the line is: y - 2 = -2(x - 5)


This isn't all, if you look at the question, we need to find where L intersects the x and y axes. We can do this by putting this equation into y = mx + x format.

y - 2 = -2x + 10
y = -2x + 12


In this format, we know that we have worked it out correctly because the gradient is still the same. Also, we know that L intersects the y axis at + 12 just by looking at the equation - when x = 0, y = 12. To get the coordinates of where it crosses the x axis we just need to swap the equation around a little.

When x = 0, y = (-2 × 0) + 12
y = 0 + 12
y = 12. L intersects the y axis at (0, 12).
y = -2x + 12
2x + y = 12
2x = 12 - y
When y = 0, 2x = 12 - 0
x = 6. L intersects the x axis at (6, 0).




January 2005



To start, we're going to work out the gradient. If you're stuck on how to do this, please visit the gradients page, here. So, taking the two points we've been given (A and B), we're going to find the gradient:

Gradient (m) of line AB = y1 - y2x1 - x2
= 7 + 13 - 5
= 8-2
= 4-1
= -4


Now we can sub all the info we have to create an equation. We're going to use the y - y1 = m(x - x1) form and then change it into the y = mx + c format wanted.

y - 7 = -4(x - 3)
y - 7 = -4x + 12
y = -4x + 19


Now we have the equation, we need the mid-point. If you're unsure on how to work out the mid-point, click here.

Mid-point of line AB = (x1 + x22 , y1 + y22)
= (3 + 52 , 7 - 12)
= (82 , 62)
= (4, 3)


Let's try subbing those coordinates into the equation we've created to see if it works and to show the mid-point lies on the line (A, B).

x + 2y = 10
4 + (2 × 3) = 10
4 + 6 = 10
10 = 10 ✓


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