KCET · Maths · Straight Lines
If the line \(6 x-7 y+8+\lambda(3 x-y+5)=0\) is parallel to \(y\)-axis, then \(\lambda\) is equal to
- A \(-7\)
- B \(-2\)
- C 7
- D 2
Answer & Solution
Correct Answer
(A) \(-7\)
Step-by-step Solution
Detailed explanation
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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