Question

If point p(3,4)are equidistance from the point a(a+b,b-a),b(a-b,a+b) prove. That 3b-4a=0

Answer 1

(a+b-3)^2+(b-a-4)^2=(a-b-3)^2+(b+a-4)^2

solving
-6a-6b+2ab-2ab+8a-8b=6b-6a-2ab+2ab-8a-8b
16a=12b
hence proved

Answer 2

Answer:

The proof is given below

Step-by-step explanation:

Given that point P(3,4)are equidistant from the point A(a+b,b-a), B(a-b,a+b)

we have to prove that 3b-4a=0

Distance between the coordinates $(x_1,y_1)\text{ and }(x_2,y_2)$ is

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

PA=PB  (Given)

⇒ $\sqrt{(a+b)-3)^2+((b-a)-4)^2}=\sqrt{((a-b)-3)^2+((a+b)-4)^2}\\\\\text{Squaring on both sides}\\\\(a+b)^2+9-6(a+b)+(b-a)^2+16-8(b-a)=(a-b)^2+9-6(a-b)+(a+b)^2+16-8(a+b)\\\\-6(a+b)-8(b-a)=-6(a-b)-8(b+a)\\\\2(a+b)=-14(a-b)\\\\-16a+12b=0\\\\3b-4a=0$

Hence proved