pure core 1 maths chap 3 - coordinate geometry of straight lines question

Find, in the form ax+by+c=0, the equation of the line with gradient 2 and y-intercept -3
I get that you're supposed to rearrange it into the form y=mx+c, but that only gives the values of a/b and c/b, not the letters by their selves.

Find, in the form ax+by+c=0, the equation of the line with gradient 2 and y-intercept -3
I get that you're supposed to rearrange it into the form y=mx+c, but that only gives the values of a/b and c/b, not the letters by their selves.

For the simplest integer coefficients in that form, a = -2, b = 1, c = 3. Any multiples of this would also work, but it's best to simplify fully.

Could you show me how you did it

All the question is asking you to do is sub in your values of m and c, then rearrange y = 2x-3 into the form 0 = 2x-y-3

You might be over complicating it

You can rearrange the equation that you had: $y = -\frac{a}{b}x -\frac{c}{b}$
Then $\frac{a}{b} = -2, \frac{c}{b} = 3$
Settting b = 1 for simplification, a = -2, c = 3

Or you can go from y = 2x - 3 and rearrange to $-2x + y + 3 = 0$

But why does a=2 when we work -a/b out as being 2?

So you're supposed to take b to equal 1 even though the question doesn't say that?

Yeah. In this situation, you can pick any(non zero) value you want for one of the variables, and the other two will change their values to match. I picked 1 because it's simple. If you don't like that, just rearrange the y = mx + c into the form you want.

Yeah. In this situation, you can pick any(non zero) value you want for one of the variables, and the other two will change their values to match. I picked 1 because it's simple. If you don't like that, just rearrange the y = mx + c into the form you want.