If the equation of the locus of point equidistant from the points ( a 1 , b 1 ) and ( a 2 , b 2 ) is ( a 1 - a 2 ) x + ( b 1 - b 2 ) y + c = 0 , then c =

Question

If the equation of the locus of point equidistant from the points $(a_{1}, b_{1})$ and $(a_{2}, b_{2})$ is $(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0$ , then $c =$

Smart Achievers · Accepted Answer

(4)

$Let α, β be the point of locus, equidistant from (a_{1}, b_{1}) and (a_{2}, b_{2}) is given by$

${(α - a_{1})}^{2} + {(β - b_{1})}^{2} = {(α - a_{2})}^{2} + {(β - b_{2})}^{2}$

$\Rightarrow$ $a_{1}^{2} + b_{1}^{2} - 2 a_{1} α - 2 b_{1} β - a_{2}^{2} - b_{2}^{2} + 2 a_{2} α + 2 b_{2} β = 0$

$\Rightarrow$ $2 (a_{2} - a_{1}) α + 2 (b_{2} - b_{1}) β + a_{1}^{2} + b_{1}^{2} - b_{2}^{2} - a_{2}^{2} = 0$

$\Rightarrow$ $(a_{2} - a_{1}) x + (b_{2} - b_{1}) y + \frac{1}{2} (a_{1}^{2} + b_{1}^{2} - a_{2}^{2} - b_{2}^{2}) = 0$

$\Rightarrow$ $c = - \frac{1}{2} [a_{1}^{2} + b_{1}^{2} - a_{2}^{2} - b_{2}^{2}] = \frac{1}{2} [a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2}]$

Solution

Q.
If the equation of the locus of point equidistant from the points $(a_{1}, b_{1})$ and $(a_{2}, b_{2})$ is $(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0$ , then $c =$

1 $a_{1}^{2} - a_{2}^{2} + b_{1}^{2} - b_{2}^{2}$

2 $\frac{1}{2} (a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2})$

3 $\sqrt{(a_{1}^{2} + b_{1}^{2} - a_{2}^{2} - b_{2}^{2})}$

4 $\frac{1}{2} (a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2})$

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Solution

Q. If the equation of the locus of point equidistant from the points (a1, b1) and (a2, b2) is (a1-a2)x+(b1-b2)y+c=0, then c=

1 a12-a22+b12-b22

2 12(a12+a22+b12+b22)

3 (a12+b12-a22-b22)