# Co-ordinate geometry help

• October 1st 2007, 06:45 AM
Chris85
Co-ordinate geometry help
Hi, what is the best way to find the equation of a straight line when given to points on the line. ie (x1,y1) and (x2,y2)

I have been finding the gradient m=y2-y1/x2-x1 and then putting into the formula y-y1 =m(x-x1) but i am doing something wrong as i keep getting different answers to my textbook. Could somebody please solve one for me so i can see how it's done?

Heres one in my book Find the equation of the line AB when A(-2,4) B(5,3)

Thanks, Chris.
• October 1st 2007, 07:19 AM
Krizalid
Find $m$

$m=\frac{3-4}{5+2}=-\frac17$

Now select one point and apply the point-slope formula.
• October 1st 2007, 07:28 AM
red_dog
The equation of a line passing through two points $M_1(x_1,y_1), \ M_2(x_2,y_2)$ is

$\displaystyle\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$

or $\displaystyle\begin{vmatrix}x & y & 1\\x_1 & y_1 & 1\\x_2 & y_2 & 1\end{vmatrix}=0$
• October 1st 2007, 07:39 AM
Chris85
Hi thanks. Yes i know the formula, it's just i am having trouble using it :o i am going wrong somewhere with it.
• October 1st 2007, 08:02 AM
topsquark
Quote:

Originally Posted by Chris85
Hi thanks. Yes i know the formula, it's just i am having trouble using it :o i am going wrong somewhere with it.

Why don't you post your solution and we'll take a look at it.

-Dan
• October 1st 2007, 09:37 AM
Chris85
OK

A=(-2,4) B=(5,3)

M= y2-y1/x2-x1 = 3-4/5-(-2) = -1/7

y-y1/y2-y1 = x-x1/x2-x1

therefore y-4/3-4 = x-(-2)/5-(-2)

Am i right so far if so what do i do next?? apologies if it's hard to understand im new and still trying to do the latex :)
• October 1st 2007, 10:03 AM
earboth
Quote:

Originally Posted by Chris85
OK

A=(-2,4) B=(5,3)

M= y2-y1/x2-x1 = 3-4/5-(-2) = -1/7

y-y1/y2-y1 = x-x1/x2-x1

therefore y-4/3-4 = x-(-2)/5-(-2)

Am i right so far if so what do i do next?? apologies if it's hard to understand im new and still trying to do the latex :)

Hello,

you calculated the slope of the line correctly.

Now you plug in the coordinates of one point into the equation of the line:

point-slope-formula, see Krizalid's post.
$\frac{y-y_1}{x-x_1} = m~\implies~\frac{y-3}{x-5}=-\frac17$ Solve for y:

$y-3 = -\frac17 \cdot(x-5)$ expand the bracket and solve for y:

$y = -\frac17 \cdot x + \frac{26}7$
• October 1st 2007, 12:16 PM
Chris85
ok, the answer in my textbook is x+7y-26=0??
• October 3rd 2007, 09:37 PM
earboth
Quote:

Originally Posted by Chris85
ok, the answer in my textbook is x+7y-26=0??

Hi,

take the result I gave you, multiply both sides by 7 and collect all terms on the LHS - then you'll get the answer which is in your book.