Question

(i) If the point P(2, 2) is equidistant from the points (a+b, b-a) and (a-b, a+b), prove that bx = ay.

(ii) If thedistances of P(x, y) from A(5, 1) and B (-1, 5) are equal then prove that 3x = 2y.

Answer 1

(i)

Distace between the points (x, y) and (a+b, b-a) & (a-b, a+b) is equal

⇒ $\sqrt{[x-(a+b){]}^2+[y-(b-a){]}^2}=\sqrt{[x-(a-b){]}^2+[y-(a+b){]}^2}$

⇒ $x^2+(a+b{)}^2-2x(a+b)+y^2+(b-a{)}^2-2y(b-a)=x^2+(a-b{)}^2-2x(a-b)+y^2+(a+b{)}^2-2y(a+b)$

⇒ $-2ax-2bx-2by+2ay=-2ax+2bx-2ay-2by$

⇒ $ay-bx=bx-ay$

⇒ $2ay=2bx$

⇒ $bx=ay$

(ii)

It is given that P is equidistant from A and B.
So, PA = PB
Using distance formula,


then

+ 25 - 10x + + 1 - 2y = + 1 + 2x + + 25 - 10y
- 12x = -8y
3x = 2y
Hence proved