# Math Help - co-ordinate geometry

1. ## co-ordinate geometry

Btw, how do I find my previous posts on here ?

OK, can someone explain. I know the gradient of a line, and i'm being asked to find the equation of the line perpendicular to it.

The gradient of the line is 1/2. So m1m2 = -1, so m2 = - 2 ?

So how do I find the equation - in the form ax + by + c = 0

The points are A (-1, -2 ) B ( 7 , 2) and C (6, 4)

please help!

2. Originally Posted by Mathematix
Btw, how do I find my previous posts on here ?
Just Click on your name and it will give you the option to view your posts.

Originally Posted by Mathematix
OK, can someone explain. I know the gradient of a line, and i'm being asked to find the equation of the line perpendicular to it.

The gradient of the line is 1/2. So m1m2 = -1, so m2 = - 2 ?

So how do I find the equation - in the form ax + by + c = 0

The points are A (-1, -2 ) B ( 7 , 2) and C (6, 4)

please help!
It looks like AB has a slope of 1/2 and BC has a slope of -2.

These two slopes are negative reciprocals of each other, meaning their product is -1. This also means that AB is perpendicular to BC.

The equation of the line containg A(-1, -2) and B(2, 2) is found by using the slope-intercept form of the linear equation ( $y=mx+b$), and either point A or B. I'll use A. Substituting, we have

$-2=\frac{1}{2}(-1)+b$
$-2=-\frac{1}{2}+b$
$-\frac{3}{2}=b$

$y=\frac{1}{2}x-\frac{3}{2}$

To translate the equation into the general form Ax + By + C = 0, first multiply each term by 2 to eliminate the denominator:

$2y=x-3$

Then, transpose (move) the 2y to the right side of the equation. Don't forget to change its sign.

$0=x-2y-3$

Finally, use the symmetric property of equality to get:

$\boxed{x-2y-3=0}$

You can follow these steps to define the equation of the perpendicular that passes through points B and C. Just remember to use the gradient (slope) of -2 and either point B or C.

Good Luck!