TS EAMCET · Maths · Straight Lines
If the normal form of the equation of a straight line \(4 x+3 y+2=0\) is \(x \cos \alpha+y \sin \alpha=p\) and its intercept form is \(\frac{x}{a}+\frac{y}{b}=1\), then \(\frac{p \sec \alpha}{a b}=\)
- A \(\frac{-1}{2}\)
- B \(\frac{3}{2}\)
- C \(\frac{-3}{2}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{-3}{2}\)
Step-by-step Solution
Detailed explanation
\begin{gathered}\text { Given, } 4 x+3 y+2=0 \Rightarrow-4 x-3 y=2 \\ \frac{-4 x}{5}-\frac{3}{5} y=\frac{2}{5} \\ \because \cos \alpha=\frac{-4}{5}, \sin \alpha=\frac{-3}{5} \text { and } P=\frac{2}{5} \\ \frac{x}{2 /-4}+\frac{y}{2 /-3}=1 \Rightarrow a=\frac{-2}{4},…
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