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Question: If the equation of the locus of a point equidistant from the points \((a_{1},b_{1})\) and \((a_{2},b...

If the equation of the locus of a point equidistant from the points (a1,b1)(a_{1},b_{1}) and (a2,b2)(a_{2},b_{2}) is

(a1−a2)x+(b1−b2)y+c=0(a_{1} - a_{2})x + (b_{1} - b_{2})y + c = 0, then the value of c is.

A

a12−a22+b12−b22a_{1}^{2} - a_{2}^{2} + b_{1}^{2} - b_{2}^{2}

B

a12+b12−a22−b22\sqrt{a_{1}^{2} + b_{1}^{2} - a_{2}^{2} - b_{2}^{2}}

C

12(a12+a22+b12+b22)\frac{1}{2}(a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2})

D

12(a22+b22−a12−b12)\frac{1}{2}(a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2})

Answer

12(a22+b22−a12−b12)\frac{1}{2}(a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2})

Explanation

Solution

Let (h,k)( h , k ) be the point on the locus, then by the given conditions

(h−a1)2+(k−b1)2=(h−a2)2+(k−b2)2\left( h - a _ { 1 } \right) ^ { 2 } + \left( k - b _ { 1 } \right) ^ { 2 } = \left( h - a _ { 2 } \right) ^ { 2 } + \left( k - b _ { 2 } \right) ^ { 2 }

⇒2h(a1−a2)+2k(b1−b2)+a22+b22−a12−b12=0\Rightarrow 2 h \left( a _ { 1 } - a _ { 2 } \right) + 2 k \left( b _ { 1 } - b _ { 2 } \right) + a _ { 2 } ^ { 2 } + b _ { 2 } ^ { 2 } - a _ { 1 } ^ { 2 } - b _ { 1 } ^ { 2 } = 0

⇒h(a1−a2)+k(b1−b2)+12(a22+b22−a12−b12)=0\Rightarrow h \left( a _ { 1 } - a _ { 2 } \right) + k \left( b _ { 1 } - b _ { 2 } \right) + \frac { 1 } { 2 } \left( a _ { 2 } ^ { 2 } + b _ { 2 } ^ { 2 } - a _ { 1 } ^ { 2 } - b _ { 1 } ^ { 2 } \right) = 0 ...(i)

Also, since (h, k) lies on the given locus, therefore

(a1−a2)h+(b1−b2)k+c=0\left( a _ { 1 } - a _ { 2 } \right) h + \left( b _ { 1 } - b _ { 2 } \right) k + c = 0 ....(ii)

Comparing (i) and (ii), we get

c=12(a22+b22−a12−b12)c = \frac { 1 } { 2 } \left( a _ { 2 } ^ { 2 } + b _ { 2 } ^ { 2 } - a _ { 1 } ^ { 2 } - b _ { 1 } ^ { 2 } \right).