JEE MAIN - Mathematics (2003 - No. 48)
If the equation of the locus of a point equidistant from the point $$\left( {{a_{1,}}{b_1}} \right)$$ and $$\left( {{a_{2,}}{b_2}} \right)$$ is
$$\left( {{a_1} - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y + c = 0$$ , then the value of $$'c'$$ is :
$$\left( {{a_1} - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y + c = 0$$ , then the value of $$'c'$$ is :
$$\sqrt {{a_1}^2 + {b_1}^2 - {a_2}^2 - {b_2}^2} $$
$${1 \over 2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right)$$
$${{a_1}^2 - {a_2}^2 + {b_1}^2 - {b_2}^2}$$
$${1 \over 2}\left( {{a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2} \right)$$.
Explanation
Since, the points $\left(a_1, b_1\right)$ and $\left(a_2, b_2\right)$ satisfy the equation. So, that
$$ \begin{aligned} & a_1\left(a_1-a_2\right)+b_1\left(b_1-b_2\right)+c=0 ~~........(1) \\\\ & \text { and } a_2\left(a_1-a_2\right)+b_2\left(b_1-b_2\right)+c=0 ~~.........(2) \\\\ & \text { On adding Eqs. (i) and (ii), we get } \\\\ & \left(a_1+a_2\right)\left(a_1-a_2\right)+\left(b_1+b_2\right)\left(b_1-b_2\right)+2 c=0 \\\\ & \Rightarrow 2 c=-\left(a_1^2-a_2^2+b_1^2-b_2^2\right) \\\\ & \Rightarrow c=\frac{1}{2}\left(a_2^2+b_2^2-a_1^2-b_1^2\right) \end{aligned} $$
$$ \begin{aligned} & a_1\left(a_1-a_2\right)+b_1\left(b_1-b_2\right)+c=0 ~~........(1) \\\\ & \text { and } a_2\left(a_1-a_2\right)+b_2\left(b_1-b_2\right)+c=0 ~~.........(2) \\\\ & \text { On adding Eqs. (i) and (ii), we get } \\\\ & \left(a_1+a_2\right)\left(a_1-a_2\right)+\left(b_1+b_2\right)\left(b_1-b_2\right)+2 c=0 \\\\ & \Rightarrow 2 c=-\left(a_1^2-a_2^2+b_1^2-b_2^2\right) \\\\ & \Rightarrow c=\frac{1}{2}\left(a_2^2+b_2^2-a_1^2-b_1^2\right) \end{aligned} $$
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