# 3 points of a triangle, 21 bits to find

• March 16th 2009, 04:12 PM
Knoll318
3 points of a triangle, 21 bits to find
Okay, the title may not have made it very clear, but I have 3 vertexes of a triangle.

$A= (3,2)$
$B= (1,-2)$
$C= (-2,1)$

I need to find a bunch of stuff, all in Standard Form (Ax+By+C=0) with integer coefficients.

I'm pretty sure on my sides, but my friend says BC is wrong.

The Sides:
AB: 2x-y+5=0
BC: x+y-5=0
AC: 5x-y-7=0

If anyone can see if I got BC wrong, could you tell me? I don't need the answers themselves, I just need to see if I'm right.

As for the medians, I got the midpoints.

D is for AB
E is for BC
F is for CD

$D= (2,0)$
$E= (-.5,-.5)$
$F= (.5,2.5)$

After that, I found the slopes. Then since I have A and B of the equation, I don't know which point to balance it out with (make it equal to 0).

So that's my first question. As these get answered I have more questions I'll tackle.

Thanks for taking the time to read this.

PS: First post!
• March 17th 2009, 12:06 AM
earboth
Quote:

Originally Posted by Knoll318
Okay, the title may not have made it very clear, but I have 3 vertexes of a triangle.

$A= (3,2)$
$B= (1,-2)$
$C= (-2,1)$

I need to find a bunch of stuff, all in Standard Form (Ax+By+C=0) with integer coefficients.

I'm pretty sure on my sides, but my friend says BC is wrong.

The Sides:
AB: 2x-y+5=0
BC: x+y-5=0
AC: 5x-y-7=0

If anyone can see if I got BC wrong, could you tell me? I don't need the answers themselves, I just need to see if I'm right.

As for the medians, I got the midpoints.

D is for AB
E is for BC
F is for CD

$D= (2,0)$
$E= (-.5,-.5)$
$F= (.5,2.5)$

After that, I found the slopes. Then since I have A and B of the equation, I don't know which point to balance it out with (make it equal to 0).

...

1. The coordinates of the given points must satisfy the equation you have determined. But that doesn't happen with your equations.

2. If you have 2 points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ then the equation of the line passing through these points is:

$\dfrac{y-y_1}{x-x_1} = \underbrace{\dfrac{y_2-y_1}{x_2-x_1}}_{\text{slope of the line}}$

3. Using this formula with A and B you'll get:

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

Transforming this equation into standard form you'll get:

$-2x+y+4=0~\vee~2x-y-4=0$

3. I've got

BC: $x+y+1=0$

AC: $-x+5y-8=0$