Coordinate Geometry

Question

Given that the points P(at^2,2at),Q(a,0) and R(a/t^2 ,(-2at)/t) where a is a positive constant and t>o,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 PR

 

Hence deduce that the perpendicular distance from O to the line PR is 2at/(1+t^2 ) .

Answer 1

It's an 11-year-old problem, so let's see how this works out.

Given P(at^2, 2at), Q(a,0), and R(a/t^2, -2at). *assume a and t are positive.

a) Show that P-A-R (the three points are collinear)

Best way to do this is to find the slope of any 2 combination of points.

m of PQ = (2at - 0)/(at^2 - a) = 2at / [a(t^2 - 1)] = 2t / (t^2 - 1)

m of QR = (-2a/t - 0)/(a/t^2 - a) = -2a / [at(1/t^2 - 1)] = -2/ [t[(1-t^2)/t^2]] = -2t^2 / [t(1-t^2)] = -2t/(1-t^2) = 2t/(t^2 - 1)

b) Find the area of OPR, where O(0,0).

Best way to find the area is to recognize that 1/2 det |x1 y1| will give the area of the triangle formed by the origin and (x1,y1) and (x2,y2) |x2 y2|

So area of OPR is 1/2 [(at^2)(-2a/t) - (a/t^2)(2at)] = 1/2 (-2ta^2 - 2a^2/t) = -a^2(t + 1/t). Finally, we'll discard the negative because area is positive. Final answer is a^2(t+1/t)

c) Find PR.

By using the distance formula: PR = sqrt [ (2at - 2a/t)^2 + (at^2 - a/t^2)^2 ]

sqrt [4a^2(t + 1/t)^2 + a^2(t^2 - 1/t^2)^2]

a * sqrt[ 4(t + 1/t)^2 + (t + 1/t)^2 * (t - 1/t)^2]

a (t + 1/t) sqrt [ 4 + (t - 1/t)^2]

d) Find the distance from O to PR.

In part b, we found the area of triangle OPR. In part c, we found the length of PR (which is the base for triangle OPR. If we take the answer from part b and then divide the answer from part c, we would get 1/2 x height of the triangle = 1/2 x (distance from O to PR).

By division, 1/2 x (distance from O to PR) = [a^2(t+1/t)] / [a (t + 1/t) sqrt [ 4 + (t - 1/t)^2]]

(distance from O to PR) = 2 * a / sqrt [ 4 + (t - 1/t)^2]]

= 2a / sqrt [ 4 + (t - 1/t)^2]]

= 2at / t * sqrt [ 4 + (t - 1/t)^2]

= 2at / sqrt [ 4t^2 + t^2 (t^2 - 2 + 1/t^2)]

= 2at / sqrt [ 4t^2 + t^4 - 2 t^2 + 1]

= 2at / sqrt [ t^4 + 2 t^2 + 1]

= 2at / sqrt [ (t^2 + 1)]

= 2at / (t^2 + 1)

That was a fun problem. :)