Question

P lies on the line y=x and Q lies on y=2x. The equation for the locus of the midpoint of PQ. if PQ=4, is

Answer 1

Answer:

25x^2−36xy+13y^2−4=0

Step-by-step explanation:

P lies on y=x, hence its coordinates are P(a,a)

Q lies on y=2x, hence its coordinates are Q(b,2b)

Given that   |PQ|=4

Hence

(a−b)^2+(a−2b)^2=4^2=16 ………………(1)

Now let M be the mid point of PQ. M has coordinates

(a+b)/2,  (a+2b)/2

To trace M’s locus, we assign x and y coordinates as :

x=(a+b)/2       ………………(2)

and

y=(a+2b)/2     ………………(3)

Solving for a and b from eqn (2) and eqn(3) above we get,

b=2y−2x         ………………(4)

a=4x−2y          ………………(5)

Substituting eqn (4) and eqn(5) into eqn(1) gives the desired locus. That’s :

16=(6x−4y)^2+(8x−6y)^2

Hence

4=(3x−2y)^2+(4x−3y)^2

Hence the equation of the locus is :

25x^2−36xy+13y^2−4=0