P (-2, 5) and Q (3, 2) are two points. Find the co-ordinates of the point R on
PQ such that PR=2QR
Answers
Step-by-step explanation:
Let say R be (x,y)
PR = √(-2-x)^2 + (5-y)^2
QR = √(3-x)^2 + (2-y)^2
Given PR = 2QR.
squaring on both sides. PR^2 = 4*QR^2.
=> (-2-x)^2 + (5-y)^2 = 4*((3-x)^2 + (2-y)^2)
=>4+x^2+4x + 25+y^2-10y =4(9+x^2-6x + 4+y^2-4y)
=> x^2+y^2+4x-10y+29 = 4(x^2+y^2-6x-4y+13)
=>x^2+y^2+4x-10y+29 = 4x^2+4y^2-24x-16y+52
=> 3x^2+3y^2-28x-6y+23= 0. ------(1) eqn.
Any point (x,y) that satisfy this equation is answer.
If R lies on line PQ, then we have to do something more. First we have to find the equation of line.
For that, follow two points method.
(y -5)/(x+2) = (5-2)/(-2-3)
(y -5)/(x+2) = -3/5
cross multiply
5y - 25 = -3x -6
5y = 19- 3x
y = (19 - 3x)/5 -------(2) eqn.
Substitute (2) eqn in (1) eqn.
Then we get a quadratic equation of x.
Find the roots and then take real roots and substitute in (2) eqn to find y.
Then the point R (x,y) will be found.
Hope you got an idea how to solve the problem.