Question

P(x, y) is a point equidistant from the points A(4, 3) and B(3, 4). Prove that x -y = 0.

Answer 1

Given, 

\(P(x, y), A(4,3)\, \& \,B(3,4)\)

\(P A=P B \text { (Equidistant) }\)

So, by distance formula we have

Distance between two points \(=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

\(P A^2=P B^2 \)

⇒ \((4-x)^2+(3-y)^2=(3-x)^2+(4-y)^2 \)

⇒ \((4)^2+(x)^2-2 \times 4 \times x+(3)^2+(y)^2-2 \times 3 \times y\) \(=(3)^2+(x)^2-2 \times 3 \times x+(4)^2+(y)^2-2 \times 4 \times y \)

⇒​​​​​​​ \( 16+x^2-8 x+9 x+y^2-6 y=9+x^2-6 x+16+y^2-8 y \)

\(-8 x+6 x-6 y+8 y=0 \)

\(-2 x+2 y=0 \)

\(-2(x-y)=0\)

\( x-y=0\).

Answer 2

\(P A^2=P B^2 \)

\(\Rightarrow(x-4)^2+(y-3)^2=(x-3)^2+(y-4)^2 \)

\(\Rightarrow x=y\) or x - y = 0