c4 question

http://filestore.aqa.org.uk/subjects/AQA-MPC4-W-MS-JAN13.PDF

http://filestore.aqa.org.uk/subjects/AQA-MPC4-QP-JAN13.PDF

in question 4 a ii
basically the mark scheme uses the y2-y2 = m(x2-x1) thing to find the equation of the lines.
basically ive never used that ive always just found the gradient and than y=mx+c stick in the gradient and an x,y value and calculate the C so ive got no idea what they did in this method. if someone can explain it in terms of the way i did it, it'd be really helpful.

How have you been finding the gradient?

The concept of y2-y1= m(x2-x1) is the most common way to find a gradient, and that's finding two coordinates of the line. If you look at the equation, divide both sides by (x2-x1) and you're finding the gradient, m.

So say you have coordinates (2,4) and (4,6). The equation would be (6-4)=m(4-2) and if you rearrange the equation, m would be 1.

That way would also work.

If you know y=mx+c, and you know the value of m and a point on the line, you can just substitute in and solve for c.

For example, say we have a line with gradient 2 passing through (5,1), we know y=2x+c.

Since we also know (5,1) is on the line, and every point on the line satisfies the equation, 5=(2*1)+c so c=3 and the equation is y=2x+3.

Hope this helps .