SOLUTION: The point P lies on the line joining A(-1,-5) and B(11,13) such that AP=(1/3)AB. (i) Find the equation of the line perpendicular to AB and passing through P. The line perpendicul

Algebra ->  Test -> SOLUTION: The point P lies on the line joining A(-1,-5) and B(11,13) such that AP=(1/3)AB. (i) Find the equation of the line perpendicular to AB and passing through P. The line perpendicul      Log On


   

Question 1172152: The point P lies on the line joining A(-1,-5) and B(11,13) such that AP=(1/3)AB.
(i) Find the equation of the line perpendicular to AB and passing through P.
The line perpendicular to AB passing through P and the line parallel to the
x-axis passing through B intersect at the point Q.
(ii) Find the coordinates of the point Q.
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
the equation of AB: slope is 3/2 and point slope y-y1=m(x-x1), m slope (x1, y1) point is
y+5=(3/2)(x+1)
y=(3/2)x-7/2
The line perpendicular to AB has slope -2/3.
AP is 4 x units or x=+3 from A and 6 y units or +1 from A.
so P is at (3, 1)
So AP has equation y-1=(-2/3) (x-3)
y=(-2/3)x+3
graph these
graph%28500%2C500%2C-15%2C15%2C-15%2C15%2C11%2C1%2C%281.5x-7%2F2%29%2C%28-2%2F3%29x%2B3%29
Point Q has y=11 and is on line y=(-2/3)x+3, so 8=(-2/3)x and x=-12
(-12, 11)