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).

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- Jul 30th 2007, 12:23 PMicebreaker09Algebra/Coordinate GeometryTriangle 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).

- Jul 30th 2007, 12:42 PMPlato
The slope of line AB is thus the altitude from C to AB is .

The slope of line AC is thus the altitude from B to AC is .

Find the intersection. - Jul 30th 2007, 03:12 PMicebreaker09What 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?

- Jul 30th 2007, 03:22 PMPlato
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. - Jul 30th 2007, 04:18 PMicebreaker09I 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..