Q78 / 180 · KCET 2013 · Mathematics

If the line $6 x-7 y+8+\lambda(3 x-y+5)=0$ is parallel to $y$-axis, then $\lambda$ is equal to

  • Correct answer
    A
    $-7$
  • B
    $-2$
  • C
    7
  • D
    2

Solution

Given line is $\begin{array}{rr} & 6 x-7 y+8+\lambda(3 x-y+5)=0 \\ \Rightarrow & (6+3 \lambda) x-(7+\lambda) y+(8+5 \lambda)=0 \\ \Rightarrow & (7+\lambda) y=(6+3 \lambda) x+(8+5 \lambda) \\ \Rightarrow & y=\frac{3(\lambda+2)}{(\lambda+7)} x+\left(\frac{8+5 \lambda}{7+\lambda}\right) \\ \therefore & \text { Slope of the line }(m)=\frac{3(\lambda+2)}{(\lambda+7)}\end{array}$ Since, line is parallel to $y$-axis. $\begin{array}{ll} \therefore & m=\infty=\frac{1}{0} \\ \Rightarrow & \frac{3(\lambda+2)}{\lambda+7}=\frac{1}{0} \\ \Rightarrow & \lambda+7=0 \\ \Rightarrow & \lambda=-7 \end{array}$

Hence, the correct answer is option: 'A'.

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