TS EAMCET · Maths · Straight Lines
\(L \equiv x \cos \alpha+y \sin \alpha-p=0\) represents a line perpendicular to the line \(x+y+1=0\). If \(p\) is positive, \(\alpha\) lies in the fourth quadrant and perpendicular distance from \((\sqrt{2}, \sqrt{2})\) to the line \(L=0\) is 5 units then \(p=\)
- A 5
- B \(\frac{5}{2}\)
- C 10
- D \(\frac{15}{2}\)
Answer & Solution
Correct Answer
(A) 5
Step-by-step Solution
Detailed explanation
Slope of line \(L=-\cot \alpha\). Slope of line \(x+y+1=0\) is \(=-1\) Since both are perpendicular to each other. \(\therefore(-\cot \alpha)(-1)=-1 \Rightarrow \tan \alpha=-1\) \(\therefore \alpha=2 \pi-\frac{\pi}{4} \quad(\because \alpha\) lies in 4th quadrant) According to…
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