SOLUTION: help please! a triangle has its vertices at (6,0), (5,3), and (-1,-4) find the equation of the line through the vertices parallel to the opposite side.

Algebra ->  Points-lines-and-rays -> SOLUTION: help please! a triangle has its vertices at (6,0), (5,3), and (-1,-4) find the equation of the line through the vertices parallel to the opposite side.      Log On


   

Question 683785: help please! a triangle has its vertices at (6,0), (5,3), and (-1,-4) find the equation of the line through the vertices parallel to the opposite side.
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
(6,0), (5,3), and (-1,-4)
Let A (6,0), B(5,3), and C(-1,-4)be the vertices of the triangle
Let P,Q be the mid points of AB & AC respectively
PQ is parallel to BC
Determine co ordinates of P & Q by mid point formula
A (6,0), B(5,3),
Let P(x3,y3)
x3=+%286%2B5%29%2F2+=+11%2F2
y3=%283%2B0%29%2F2+=+3%2F2
p=((11/2),(3/2))
Similarly find co ordinates of Q the mid point of AC
Q=(x4,y4)
A (6,0), C(-1,-4)
x4=+%28-1%2B6%29%2F2=%285%2F2%29
y4=%28-4%2B0%29%2F2=+-2
Q=((5/2),(-2))
The equation of PQ
use formula %28%28y-y1%29%2F%28y1-y2%29%29=%28%28x-x1%29%2F%28x1-x2%29%29
p=((11/2),(3/2)), Q=((5/2),(-2))
equation of PQ ->
%28y-3%2F2%29%2F%287%2F2%29+=+%28x-%2811%2F2%29%29%2F3
3%28y-%283%2F2%29%29=%287%2F2%29%28x-%2811%2F2%29%29
3y-%289%2F2%29=7x%2F2-%2877%2F4%29
multiply equation by 4
12y-18=14x-77
14x-12y=59 is the equation of PQ