TS EAMCET · Maths · Straight Lines
The equation of a given straight line is \(\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=\gamma\). If the equation of the line perpendicular to the given line and passing through \((\alpha, \beta)\) is \(\frac{x}{a}+\frac{y}{b}=1\), then \(\frac{b}{a}\) is equal to
- A \(\tan \theta\)
- B \(-\tan \theta\)
- C \(\cot \theta\)
- D \(-\cot \theta\)
Answer & Solution
Correct Answer
(C) \(\cot \theta\)
Step-by-step Solution
Detailed explanation
Given equation of line is \(\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=\gamma\) Slope of Eq. (i) is \(m=\tan \theta\) So, slope of a line perpendicular to Eq. (i) is \((-\cot \theta)\). According to the question, equation of required line is…
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