# Thread: Finding the equation of a straight line...

1. ## Finding the equation of a straight line...

Ok I think I've worked this out incorrectly so if someone could point me in the right direction that would be great.....

The question is find the equation of a line that passes through (-1,5) and (2,-4)
I know that
y=mx+c to find M = change in y value divide by change in x value=

2- (-1)/5-(-4) = 3/9 =1/3

so y = 1/3x=c

so 5=1/3(-1)+c
5= -1/3+c
5+-1/3 = c
4 2/3 = c so
y= 1/3x+ 14/3

You see why I need help right?

2. The question is find the equation of a line that passes through (-1,5) and (2,-4).

putting the x and y values into y = mx + c we get:
(1) 5 = -m + c
(2) -4 = 2m + c

(1) - (2) we get:
5 - (-4) = -m - 2m + c - c
9 = -3m
m = -3

Putting the value m=-3 into our (1)
5 = 3 + c

therefore c = 2

So the equation for the line is y = -3x + 2
I hope u can understand that.

edit:
What u did what "2- (-1)/5-(-4) = 3/9 =1/3"
which is the change in x divided by the change in Y.

Y has gone from 5 to -4 so the change in Y is -9, X has gone from -1 to 2 so the change in X is 3.
-9/3 = -3.

3. Originally Posted by Geo
Ok I think I've worked this out incorrectly so if someone could point me in the right direction that would be great.....

The question is find the equation of a line that passes through (-1,5) and (2,-4)
You denote the equation of a line as $y = mx + c$. Finding the gradient of the line:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - -4}{-1 - 2} = \frac{9}{-3} = -3$

Having found the gradient, we can find the equation of a line by either using the gradient-intercept form or point-gradient form:

$y = mx + c$ : choose a given point...I'll use (2, -4)

Thus, $-4 = -3 \times {2} + c$

$-4 = -6 + c$

$2 = c$

Therefore, the equation of the line is $y = -3x + 2$

* Using Point - Gradient method:

$y - y_1 = m(x - x_1)$ : again, I'll use the point (2, -4)

$y - -4 = -3(x - 2)$

$y + 4 = -3x + 6$

$y = -3x + 2$

Trust this helps.