AP EAMCET · Maths · Straight Lines
The vertex \(A\) of a triangle lies on the lines \(x+y=1\) and \(2 x+3 y=6\). If the orthocentre of the triangle is \(O\left(\frac{3}{7}, \frac{22}{7}\right)\), then the equation of \(O A\) in the normal form is
- A \(x \cos \alpha+y \sin \alpha=7 ; \alpha=\tan ^{-1} \frac{1}{7}\)
- B \(x \cos \alpha+y \sin \alpha=\frac{13}{\sqrt{17}} ; \alpha=\tan ^{-1}\left(\frac{1}{4}\right)\)
- C \(x \cos \alpha+y \sin \alpha=\frac{13}{4} ; \alpha=\tan ^{-1}\left(\frac{13}{\sqrt{17}}\right)\)
- D \(x \cos \alpha+y \sin \alpha=\frac{13}{\sqrt{17}} ; \alpha=\tan ^{-1}\)
Answer & Solution
Correct Answer
(D) \(x \cos \alpha+y \sin \alpha=\frac{13}{\sqrt{17}} ; \alpha=\tan ^{-1}\)
Step-by-step Solution
Detailed explanation
Vertex \(A\) is point of intersection of the lines \(x+y=1\) and \(2 x+3 y=6\) Point of intersection \(=(-3,4)\) So, \(A(-3,4)\) and \(O\left(\frac{3}{7}, \frac{22}{7}\right)\) Equation of line passing through two points…
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