MHT CET · Maths · Straight Lines
The co-ordinates of point on the line \(x+y+3=0\), whose distance from the line \(x+2 y+2=0\) is \(\sqrt{5}\) units, are
- A (-1,-4)
- B (1,-4)
- C (-1,4)
- D (1,4)
Answer & Solution
Correct Answer
(B) (1,-4)
Step-by-step Solution
Detailed explanation
Any point on the line \(x+y+3=0\) can be taken as \((k,-3-k)\) Now, its distance from \(x+2 y+2=0\) is
\(\begin{aligned}
& \frac{|k+2 \times(-3-k)+2|}{\sqrt{1^2+2^2}}=\sqrt{5} \\
& \Rightarrow 1-k-41=5 \\
& \Rightarrow-k-4= \pm 5 \\
& \Rightarrow k=1,-9
\end{aligned}\)
for \(k=1\) the point is \((1,-4)\)
\(\begin{aligned}
& \frac{|k+2 \times(-3-k)+2|}{\sqrt{1^2+2^2}}=\sqrt{5} \\
& \Rightarrow 1-k-41=5 \\
& \Rightarrow-k-4= \pm 5 \\
& \Rightarrow k=1,-9
\end{aligned}\)
for \(k=1\) the point is \((1,-4)\)
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