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Straight Lines and Pair of Straight Lines question

2005 · Shift 0 · Q113
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Straight Lines and Pair of Straight Lines question

2005 · Shift 0 · Q113

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
If non zero numbers a,b,ca, b, ca,b,c are in H.P.,H.P.,H.P., then the straight line xa+yb+1c=0{x \over a} + {y \over b} + {1 \over c} = 0ax​+by​+c1​=0 always passes through a fixed point. That point is :
  1. A
    (−1,2)(-1,2)(−1,2)
  2. B
    (−1,−2)(-1, -2)(−1,−2)
  3. C
    (1,−2)(1, -2)(1,−2)
  4. D
    (1,−12)\left( {1, - {1 \over 2}} \right)(1,−21​)
View written solutionFree

Correct answer: C

  1. Since a,b,ca,b,ca,b,c are in H.P. and non-zero, their reciprocals are in A.P.

    Let u=1a,v=1b,w=1c.u=\frac{1}{a},\quad v=\frac{1}{b},\quad w=\frac{1}{c}.u=a1​,v=b1​,w=c1​. Then u,v,wu,v,wu,v,w are in A.P., so 2v=u+w.2v=u+w.2v=u+w.

  2. The given line is xa+yb+1c=0.\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0.ax​+by​+c1​=0. In terms of u,v,wu,v,wu,v,w, this becomes xu+yv+w=0.xu+yv+w=0.xu+yv+w=0.

  3. Using the A.P. relation w=2v−uw=2v-uw=2v−u, substitute into the line equation: xu+yv+(2v−u)=0.xu+yv+(2v-u)=0.xu+yv+(2v−u)=0. Rearranging, u(x−1)+v(y+2)=0.u(x-1)+v(y+2)=0.u(x−1)+v(y+2)=0.

  4. For the line to pass through a fixed point (x,y)(x,y)(x,y) for all admissible u,vu,vu,v, the above must hold identically. Since uuu and vvv can vary, we require x−1=0,y+2=0.x-1=0,\quad y+2=0.x−1=0,y+2=0. Hence, x=1,y=−2.x=1,\quad y=-2.x=1,y=−2.

  5. Therefore the fixed point is (1,−2).\boxed{(1,-2)}.(1,−2)​.

  6. Checking options:

  • A: (−1,2)(-1,2)(−1,2) — incorrect
  • B: (−1,−2)(-1,-2)(−1,−2) — incorrect
  • C: (1,−2)(1,-2)(1,−2) — correct
  • D: (1,−12)\left(1,-\frac12\right)(1,−21​) — incorrect

So the correct option is C.

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