No, c is the y-intercept. It should be
where x = 2, y = 4, m = gradient = 3
I need to find the equation for a straight line that goes through point (2,4) and has a gradient of 3
I know the equation is y=mx+c but I don't have c. If I treat the point I was given as c (c=4) and add 2 to x will it be right?
y=3(x+2)+4
Or are the brackets messing it up? It doesn't look like it would be right. Could it be;
y=3x+6
That also doesn't seem right. I don't know.
Thanks in advance!
Another way to find the eqn. of a line when you don't have the y-intercept is to use the eqn. where is a point on the line and m is the gradient.
e.g.
I prefer to use y=mx+c when I have the y-intercept, but y-b=m(x-a) when I don't.
Also the latter eqn. is used more in the Higher course in Scotland.
You can check that this is incorrect by setting x= 2: y= 3(2+ 2)+ 4= 12+ 4= 16, not 4. The formula you were trying to remember is . You might aso recall that the slope of a line between two points and is [tex]\frac{y_1- y_2}{x_1- x_2}[/itex]. If you know that the slope is 3 and that one point on the line is (2, 4), then that means that for a "generic" point, (x, y), so that so that as I said above.
Another way of doing this problem is to note that the slope is "rise over run". With x going from 2 to 4, the "run" is -4 so the "rise" is 3(-4)= -12. y goes from 2 down to 2- 12= -10.
Again, if x= 2, y= 3(2)+ 6= 12, not 4 so that can't be right.Or are the brackets messing it up? It doesn't look like it would be right. Could it be;
y=3x+6
That also doesn't seem right. I don't know.
Thanks in advance!