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Question: If the point (*x, y*) be equidistant from the points \((a + b,b - a)\) and \((a - b,a + b)\), then...

If the point (x, y) be equidistant from the points (a+b,b−a)(a + b,b - a) and (a−b,a+b)(a - b,a + b), then

A

ax+by=0ax + by = 0

B

ax−by=0ax - by = 0

C

bx+ay=0bx + ay = 0

D

bx−ay=0bx - ay = 0

Answer

bx−ay=0bx - ay = 0

Explanation

Solution

Let points P(x,y)P(x,y), A(a+b,b−a),(a + b,b - a), B(a−b,a+b)B(a - b,a + b).

According to Question, PA=PBPA = PB, i.e., PA2=PB2PA^{2} = PB^{2}

⇒(a+b−x)2+(b−a−y)2=(a−b−x)2+(a+b−y)2⇒(a+b)2+x2−2x(a+b)+(b−a)2+y2−2y(b−a)\Rightarrow (a + b - x)^{2} + (b - a - y)^{2} = (a - b - x)^{2} + (a + b - y)^{2} \Rightarrow (a + b)^{2} + x^{2} - 2x(a + b) + (b - a)^{2} + y^{2} - 2y(b - a)

=(a−b)2+x2−2x(a−b)+(a+b)2+y2−2y(a+b)(a - b)^{2} + x^{2} - 2x(a - b) + (a + b)^{2} + y^{2} - 2y(a + b)

⇒2x(a−b−a−b)=2y(b−a−a−b)\Rightarrow 2x(a - b - a - b) = 2y(b - a - a - b)

⇒−4bx=−4ay⇒bx−ay=0\Rightarrow - 4bx = - 4ay \Rightarrow bx - ay = 0