MHT CET · Maths · Straight Lines
The equation of the line perpendicular to \(2 x-3 y+5=0\) and making an intercept 3 with positive Y-axis is
- A \(3 x+2 y-6=0\)
- B \(3 x+2 y+6=0\)
- C \(3 x+2 y-7=0\)
- D \(3 x+2 y-12=0\)
Answer & Solution
Correct Answer
(A) \(3 x+2 y-6=0\)
Step-by-step Solution
Detailed explanation
Let the line be \(3 x+2 y+\lambda=0\)
Putting \(x=0, y=\frac{-\lambda}{2}=3\) [given]
\(\Rightarrow \lambda=-6\)
Hence the required line is \(3 x+2 y-6=0\)
Putting \(x=0, y=\frac{-\lambda}{2}=3\) [given]
\(\Rightarrow \lambda=-6\)
Hence the required line is \(3 x+2 y-6=0\)
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