MHT CET · Maths · Straight Lines
The co-ordinates of the foot of perpendicular, drawn from the point \((-2,3)\) on the line \(3 x-y-1=0\) are
- A \((-1,2)\)
- B \((1,-2)\)
- C \((-1,-2)\)
- D \((1,2)\)
Answer & Solution
Correct Answer
(D) \((1,2)\)
Step-by-step Solution
Detailed explanation
Let \((h, k)\) be the required point on the line \(3 x-y-1=0\).
\(\therefore \quad 3 \mathrm{~h}-\mathrm{k}-1=0...(i)\)
Slope of \(3 x-y-1=0\) is 3 .
Line passing through \((\mathrm{h}, \mathrm{k})\) and \((-2,3)\) is perpendicular to \(3 x-y-1=0\).
\(\begin{aligned}
& \therefore \quad \frac{k-3}{h+2} \times 3=-1 \\
& \Rightarrow \mathrm{~h}+3 \mathrm{k}=7...(ii)
\end{aligned}\)
Solving (i) and (ii), we get \(\mathrm{h}=1\) and \(\mathrm{k}=2\)
\(\therefore \quad 3 \mathrm{~h}-\mathrm{k}-1=0...(i)\)
Slope of \(3 x-y-1=0\) is 3 .
Line passing through \((\mathrm{h}, \mathrm{k})\) and \((-2,3)\) is perpendicular to \(3 x-y-1=0\).
\(\begin{aligned}
& \therefore \quad \frac{k-3}{h+2} \times 3=-1 \\
& \Rightarrow \mathrm{~h}+3 \mathrm{k}=7...(ii)
\end{aligned}\)
Solving (i) and (ii), we get \(\mathrm{h}=1\) and \(\mathrm{k}=2\)
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