Graph; have gradient, need y intercept

• Jul 2nd 2012, 08:11 PM
Graph; have gradient, need y intercept
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.

• Jul 2nd 2012, 08:15 PM
richard1234
Re: Graph; have gradient, need y intercept
No, c is the y-intercept. It should be

$\displaystyle y = mx + c$ where x = 2, y = 4, m = gradient = 3

$\displaystyle 4 = 2(3) + c \Rightarrow 4 = 6+c$

$\displaystyle c = -2$
• Jul 10th 2012, 05:06 AM
robbiecee2
Re: Graph; have gradient, need y intercept
Another way to find the eqn. of a line when you don't have the y-intercept is to use the eqn. $\displaystyle y-b=m(x-a)$ where $\displaystyle (a,b)$ is a point on the line and m is the gradient.
e.g.
$\displaystyle m=3, pt (2,4) so,$
$\displaystyle a=2, b=4$
$\displaystyle y-b=m(x-a)$
$\displaystyle y-4=3(x-2)$
$\displaystyle y-4=3x-6$
$\displaystyle y=3x-2$

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.
• Jul 10th 2012, 05:47 AM
HallsofIvy
Re: Graph; have gradient, need y intercept
Quote:

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

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 $\displaystyle y= m(x- x_0)+ y_0$. You might aso recall that the slope of a line between two points $\displaystyle (x_1, y_1)$ and $\displaystyle (x_2, y_2)$ 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), $\displaystyle 3= \frac{y- 4}{x- 2}$ so that $\displaystyle y- 4= 3(x- 2)$ so that $\displaystyle y= 3(x- 2)+ 4$ 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.
Quote:

Or are the brackets messing it up? It doesn't look like it would be right. Could it be;

y=3x+6
Again, if x= 2, y= 3(2)+ 6= 12, not 4 so that can't be right.

Quote:

That also doesn't seem right. I don't know.