WBJEE · Maths · Straight Lines
A line passing through the point of intersection of \(x+y=4\) and \(x-y=2\) makes an angle \(\tan ^{-1}\left(\frac{3}{4}\right)\) with the \(x\) -axis. It intersects the parabola \(y^{2}=4(x-3)\) at points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) respectively. Then, \(\left|x_{1}-x_{2}\right|\) is equal to
- A \(\frac{16}{9}\)
- B \(\frac{32}{9}\)
- C \(\frac{40}{9}\)
- D \(\frac{80}{9}\)
Answer & Solution
Correct Answer
(B) \(\frac{32}{9}\)
Step-by-step Solution
Detailed explanation
Given lines are \[ x+y=4 \text { and } x-y=2 \] On solving these lines, we get \(x=3\) and \(y=1\) Now, the equation of line which passes through the intersection point (3, 1) having slope \(\theta=\tan ^{-1}\left(\frac{3}{4}\right)\) is \((y-1)=\frac{3}{4}(x-3)\)…
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