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Straight Lines and Pair of Straight Lines question

2003 · Shift 0 · Q112
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Straight Lines and Pair of Straight Lines question

2003 · Shift 0 · Q112

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
If the equation of the locus of a point equidistant from the point (a1,b1)\left( {{a_{1,}}{b_1}} \right)(a1,​b1​) and (a2,b2)\left( {{a_{2,}}{b_2}} \right)(a2,​b2​) 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(a1​−a2​)x+(b1​−b2​)y+c=0, then the value of ′c′'c'′c′ is :
  1. A
    a12+b12−a22−b22\sqrt {{a_1}^2 + {b_1}^2 - {a_2}^2 - {b_2}^2}a1​2+b1​2−a2​2−b2​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)21​(a2​2+b2​2−a1​2−b1​2)
  3. C
    a12−a22+b12−b22{{a_1}^2 - {a_2}^2 + {b_1}^2 - {b_2}^2}a1​2−a2​2+b1​2−b2​2
  4. D
    12(a12+a22+b12+b22){1 \over 2}\left( {{a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2} \right)21​(a1​2+a2​2+b1​2+b2​2).
View written solutionFree

Correct answer: B

  1. Let the moving point be P(x,y)P(x,y)P(x,y).

  2. Since PPP is equidistant from A(a1,b1)A(a_1,b_1)A(a1​,b1​) and B(a2,b2)B(a_2,b_2)B(a2​,b2​), we must have PA=PBPA=PBPA=PB so (x−a1)2+(y−b1)2=(x−a2)2+(y−b2)2.\sqrt{(x-a_1)^2+(y-b_1)^2}=\sqrt{(x-a_2)^2+(y-b_2)^2}.(x−a1​)2+(y−b1​)2​=(x−a2​)2+(y−b2​)2​.

  3. Squaring both sides, (x−a1)2+(y−b1)2=(x−a2)2+(y−b2)2.(x-a_1)^2+(y-b_1)^2=(x-a_2)^2+(y-b_2)^2.(x−a1​)2+(y−b1​)2=(x−a2​)2+(y−b2​)2.

  4. Expand both sides: x2−2a1x+a12+y2−2b1y+b12=x2−2a2x+a22+y2−2b2y+b22.x^2-2a_1x+a_1^2+y^2-2b_1y+b_1^2=x^2-2a_2x+a_2^2+y^2-2b_2y+b_2^2.x2−2a1​x+a12​+y2−2b1​y+b12​=x2−2a2​x+a22​+y2−2b2​y+b22​.

  5. Cancel x2x^2x2 and y2y^2y2: −2a1x−2b1y+a12+b12=−2a2x−2b2y+a22+b22.-2a_1x-2b_1y+a_1^2+b_1^2=-2a_2x-2b_2y+a_2^2+b_2^2.−2a1​x−2b1​y+a12​+b12​=−2a2​x−2b2​y+a22​+b22​.

  6. Rearranging, 2(a2−a1)x+2(b2−b1)y+(a12+b12−a22−b22)=0.2(a_2-a_1)x+2(b_2-b_1)y+(a_1^2+b_1^2-a_2^2-b_2^2)=0.2(a2​−a1​)x+2(b2​−b1​)y+(a12​+b12​−a22​−b22​)=0.

  7. Divide throughout by −2-2−2 to match the given form: (a1−a2)x+(b1−b2)y+12(a22+b22−a12−b12)=0.(a_1-a_2)x+(b_1-b_2)y+\frac{1}{2}(a_2^2+b_2^2-a_1^2-b_1^2)=0.(a1​−a2​)x+(b1​−b2​)y+21​(a22​+b22​−a12​−b12​)=0.

  8. Comparing with (a1−a2)x+(b1−b2)y+c=0,(a_1-a_2)x+(b_1-b_2)y+c=0,(a1​−a2​)x+(b1​−b2​)y+c=0, we get c=12(a22+b22−a12−b12).c=\frac{1}{2}(a_2^2+b_2^2-a_1^2-b_1^2).c=21​(a22​+b22​−a12​−b12​).

  9. Hence the correct option is: B.\boxed{\text{B}}.B​.

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