If the equation of the locus of a point equidistant from the points (a1, b1) and (a2, b2) is (a1 − b2) x + (a1 − b2) y + c = 0, then the value of ‘c’ is

Question

If the equation of the locus of a point equidistant from the points (a1, b1) and (a2, b2) is (a1 − b2) x + (a1 − b2) y + c = 0, then the value of ‘c’ is (A) (1/2)(a22+b22-a21-b21) (B) a21+a22-b21-b22 (C) (1/2)(a21+a22-b21-b22) (D) √a21+b22-a22-b22

Tardigrade Admin · Accepted Answer

Let p (x, y) (x â a1)2 + (y â b1)2 = (x â a2)2 + (y â b2)2 (a1-a2)x+(b1-b2)y+(1/2)(b22-b21+a22-a21)=0. Hence, (A) is the correct answer.

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

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

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

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

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

Let $p (x, y)$
$(x - a_{1})^{2} + (y - b_{1})^{2} = (x - a_{2})^{2} + (y - b_{2})^{2}$
$(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + \frac{1}{2} (b_{2}^{2} - b_{1}^{2} + a_{2}^{2} - a_{1}^{2}) = 0.$
Hence, (A) is the correct answer.

Q. If the equation of the locus of a point equidistant from the points (a1​,b1​) and (a2​,b2​) is (a1​−b2​)x+(a1​−b2​)y+c=0, then the value of ‘c’ is

21​(a22​+b22​−a12​−b12​)

a12​+a22​−b12​−b22​

21​(a12​+a22​−b12​−b22​)

a12​+b22​−a22​−b22​​