SOLUTION: A triangle has vertices at A(2, -1), B(-1, 1) and C(6,3). Find the length of its altitude drawn to the side AC.

Algebra ->  Triangles -> SOLUTION: A triangle has vertices at A(2, -1), B(-1, 1) and C(6,3). Find the length of its altitude drawn to the side AC.      Log On


   

Question 1008239: A triangle has vertices at A(2, -1), B(-1, 1) and C(6,3). Find the length of its altitude drawn to the side AC.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The slope of AC is
,
and the equation of line AC in in point-slope form is
y-y%5BA%5D=red%281%29%28x-x%5BA%5D%29-->y-%28-1%29=red%281%29%28x-2%29
We can convert to the slope-intercept form by solving for y ;
y-%28-1%29=red%281%29%28x-2%29-->y%2B1=x-2-->y=x-2-1-->y=x-3 .
The altitude to AC is the perpendicular segment from B to line AC ,
so its length is the distance from point B to line AC.
There is somewhere an ugly formula to calculate distance from a point to a line,
so we could plug numbers into that formula and be done.

Otherwise, since the altitude to AC is perpendicular to AC ,
the slope of that altitude is -1%2Fred%281%29=-1 .
The altitude to AC is part of a line
with slope -1 that contains point B%28-1%2C1%29 .
The equation of that line in point-slope form is
y-y%5BB%5D=-1%28x-x%5BB%5D%29-->y-1=-1%28x-%28-1%29%29 .
We can convert to the slope-intercept form by solving for y ;
y-1=-1%28x-%28-1%29%29-->y-1=-1%28x%2B1%29-->y-1=-x-1%29-->y=-x .
We can find point D at the intersection of the two perpendicular lines
system%28y=x-3%2Cy=-x%29-->system%28-x=x-3%2Cy=-x%29-->system%28-2x=-3%2Cy=-x%29-->system%28x=%28-3%29%2F%28-2%29%2Cy=-x%29-->system%28x=1.5%2Cy=-x%29-->system%28x=1.5%2Cy=-1.5%29 .
So the altitude is BD with D%281.5%2C-1.5%29 ,
and the length of altitude BD is