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