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QJEEM03Ma8d43: JEE Main 2003 Mathematics - Class 11 JEEM03Ma8d43 Solution: If the equation of the locus of a point equidistant from the…

If the equation of the locus of a point equidistant from the point (a1,b1)\left( {{a_{1,}}{b_1}} \right) and (a2,b2)\left( {{a_{2,}}{b_2}} \right) is
(a1−a2)x+(b1−b2)y+c=0\left( {{a_1} - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y + c = 0 , then the value of ′c′'c' is :

Multiple Choice Options

  • A a12+b12−a22−b22\sqrt {{a_1}^2 + {b_1}^2 - {a_2}^2 - {b_2}^2}
  • B 12(a22+b22−a12−b12){1 \over 2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right)
  • C a12−a22+b12−b22{{a_1}^2 - {a_2}^2 + {b_1}^2 - {b_2}^2}
  • D 12(a12+a22+b12+b22){1 \over 2}\left( {{a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2} \right) .

Step-by-Step Verified Solution

Since, the points (a1,b1)\left(a_1, b_1\right) and (a2,b2)\left(a_2, b_2\right) satisfy the equation. So, that

amp;a1(a1−a2)+b1(b1−b2)+c=0  ........(1)amp; and a2(a1−a2)+b2(b1−b2)+c=0  .........(2)amp; On adding Eqs. (i) and (ii), we get amp;(a1+a2)(a1−a2)+(b1+b2)(b1−b2)+2c=0amp;⇒2c=−(a12−a22+b12−b22)amp;⇒c=12(a22+b22−a12−b12)\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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