1. ## Algebra/Coordinate Geometry

Triangle ABC has vertices A (-5,-2), B (7,-5), and C (3,1). Find the coordinates of the intersection of the three altitudes algebraically. (I know how to graph it, I don't know how to do it algebraically).

2. The slope of line AB is $\displaystyle \frac{-1}{4}$ thus the altitude from C to AB is $\displaystyle y = 4\left( {x - 3} \right) + 1$.

The slope of line AC is $\displaystyle \frac{3}{8}$ thus the altitude from B to AC is $\displaystyle y = \frac{-8}{3}\left( {x - 7} \right) - 5$.

Find the intersection.

3. What form are those equations in? So far I've only learned standard, point-slope, and slope intercept. It seems like you combined slope-intercept and point slope... I know to find the intersection of two equations you have to simplify then use systems, and I tried playing around with your equations, but I got a fraction....
Could you help more?

4. First: The altitudes are all concurrent.
So all we need is to find the intersection of any two.
Also if a line has a non-zero slope of m then a perpendicular line has slope (-1/m).
The two lines I gave you are in point-slope form.

5. I think you misunderstood my question... I know how you came up with the equations (and now I know what form they are in) and I understand that by finding two intersection points, you find all three.

What I'm having a problem with is solving them--here's what I tried:
y= 4 (x-3) + 1
y= 4x - 13 + 1
y= 4x - 11

y= -8/3(x-7)-5
y= -8/3x + 56/3 -5 multiply everything by three (to get rid of the fraction)
3y= -8x + 56 - 15
3y = -8x + 41

And then I would use systems (for 3y = -8x + 41 and y= 4x - 11) to solve for x and y. But I get fractions/decimals..