Question

P (-2, 5) and Q (3, 2) are two points. Find the co-ordinates of the point R on

PQ such that PR=2QR​

Answer 1

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.