# Coordinate Geometry

• Sep 12th 2008, 02:27 PM
Rajivb
Coordinate Geometry
Given the points P(at2 , 2at), Q(a,0) and R (a/t2 , -2a/t) where a is a positive constant and t > 0, show that P,Q and R are collinear. Find, in terms of a and t,

(a) the area of the triangle OPR where O is the Origin.
(b) the length of PR.

Hence, deduce that the perpendicular distance from O to the line PR is 2at/1+t2.

(Wondering)
• Sep 12th 2008, 04:26 PM
ticbol
Quote:

Originally Posted by Rajivb
Given the points P(at2 , 2at), Q(a,0) and R (a/t2 , -2a/t) where a is a positive constant and t > 0, show that P,Q and R are collinear. Find, in terms of a and t,

(a) the area of the triangle OPR where O is the Origin.
(b) the length of PR.

Hence, deduce that the perpendicular distance from O to the line PR is 2at/1+t2.

(Wondering)

There is nothing much difficult here except thet it needs too much algebra. Otherwise it is straightforward.

For P,Q,R to be collinear, just show that the slope of PQ is the same as the slope of Qr.
Both slopes were solved to be 2t/(t^2 -1), so P, Q,r are collinear.

The area of triangle OPR can be solved by using the Heron's formula since the coordinates of O,P,R are known. Too much algebra.

The length of PR can be solved by the distance formula
d = sqrt[(x1 -x2)^2 +(y1 -y2)^2]

To get the perpendicular from O to PR, apply the formula for a triangle
Area = (1/2)(base)(perpendicular height of the apex from the base)
The base here is PR.
The perpendicular height of the apex, O, from PR is what you are looking for.
Area is the one you get in one of the parts above.

If you have time to do the algebra, go ahead.