MHT CET · Maths · Straight Lines
If the length of perpendicular drawn from the point \((4,1)\) on the line \(3 x-4 y+k=0\)
is 2 units, then the values of \(\mathrm{k}\) are
- A \(2,-18\)
- B \(-2,-18\)
- C \(-2,1\)
- D \(-2,18\)
Answer & Solution
Correct Answer
(A) \(2,-18\)
Step-by-step Solution
Detailed explanation
We have, point \((4,1)\) and \(3 x-4 y+k=0\)
\(\begin{array}{l}
\therefore\left|\frac{3 \times 4-4 \times 1+k}{\sqrt{9+16}}\right|=2 \Rightarrow\left|\frac{12-4+k}{\sqrt{25}}\right|=2 \\
8+k \mid=10 \Rightarrow \pm(8+k)=10 \\
8+k=10 \text { or }-8-k=10 \Rightarrow k=2 \text { or } k=-18
\end{array}\)
\(\begin{array}{l}
\therefore\left|\frac{3 \times 4-4 \times 1+k}{\sqrt{9+16}}\right|=2 \Rightarrow\left|\frac{12-4+k}{\sqrt{25}}\right|=2 \\
8+k \mid=10 \Rightarrow \pm(8+k)=10 \\
8+k=10 \text { or }-8-k=10 \Rightarrow k=2 \text { or } k=-18
\end{array}\)
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