The locus of the point equidistant from the points (a+b,a−b) and (a−b,a+b) is
A
bx−ay=0
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B
bx+ay=0
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C
ax−by=0
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D
x−y=0
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Solution
The correct option is Cx−y=0 Let the point be P(x,y) Then √(x−(a+b))2+(y−(a−b))2=√(x−(a−b))2+(y−(a+b))2 (x−(a+b))2+(y−(a−b))2=(x−(a−b))2+(y−(a+b))2 (x−(a+b))2−(x−(a−b))2=(y−(a+b))2−(y−(a−b))2 (2x−2a)(a−b−(a+b))=(2y−2a)(a−b−(a+b)) 2(x−a)(−2b)=2(y−a)(−2b) x−a=y−a or x−y=0 is the required equation.