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Straight Lines and Pair of Straight Lines question

2019 · 8 Apr · Shift 2 · Q37
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Straight Lines and Pair of Straight Lines question

2019 · 8 Apr · Shift 2 · Q37

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
If the system of linear equations x – 2y + kz = 1 2x + y + z = 2 3x – y – kz = 3 has a solution (x,y,z), z eee 0, then (x,y) lies on the straight line whose equation is :
  1. A
    4x – 3y – 4 = 0
  2. B
    3x – 4y – 1 = 0
  3. C
    4x – 3y – 1 = 0
  4. D
    3x – 4y – 4 = 0
View written solutionFree

Correct answer: A

  1. Given system

We have x−2y+kz=1...(1)x-2y+kz=1 \quad ...(1)x−2y+kz=1...(1) 2x+y+z=2...(2)2x+y+z=2 \quad ...(2)2x+y+z=2...(2) 3x−y−kz=3...(3)3x-y-kz=3 \quad ...(3)3x−y−kz=3...(3) with the condition that the system has a solution and z≠0z\ne 0z=0.

We need the line on which (x,y)(x,y)(x,y) lies.

  1. Eliminate kzkzkz using equations (1) and (3)

Add (1) and (3): (x−2y+kz)+(3x−y−kz)=1+3\big(x-2y+kz\big)+\big(3x-y-kz\big)=1+3(x−2y+kz)+(3x−y−kz)=1+3 4x−3y=44x-3y=44x−3y=4 So, 4x−3y−4=04x-3y-4=04x−3y−4=0

This already gives a straight line for (x,y)(x,y)(x,y).

  1. Why is the condition z≠0z\ne 0z=0 relevant?

Let us verify consistency with the remaining equation.

From (2): z=2−2x−yz=2-2x-yz=2−2x−y

From (1): kz=1−x+2ykz=1-x+2ykz=1−x+2y

From (3): −kz=3−3x+y⇒kz=−3+3x−y-kz=3-3x+y \Rightarrow kz=-3+3x-y−kz=3−3x+y⇒kz=−3+3x−y

Equating the two expressions for kzkzkz: 1−x+2y=−3+3x−y1-x+2y=-3+3x-y1−x+2y=−3+3x−y 4=4x−3y4=4x-3y4=4x−3y 4x−3y−4=04x-3y-4=04x−3y−4=0 Again the same line is obtained.

The condition z≠0z\ne 0z=0 ensures that some value of kkk can be chosen from k=1−x+2yzk=\dfrac{1-x+2y}{z}k=z1−x+2y​, but the locus of (x,y)(x,y)(x,y) is still determined by 4x−3y−4=0.4x-3y-4=0.4x−3y−4=0.

  1. Check options
  • A: 4x−3y−4=04x-3y-4=04x−3y−4=0 ✅
  • B: 3x−4y−1=03x-4y-1=03x−4y−1=0 ❌
  • C: 4x−3y−1=04x-3y-1=04x−3y−1=0 ❌
  • D: 3x−4y−4=03x-4y-4=03x−4y−4=0 ❌
  1. Final answer

The required straight line is 4x−3y−4=0\boxed{4x-3y-4=0}4x−3y−4=0​ So the correct option is A.

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