Question 6

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).