AP EAMCET · Maths · Straight Lines
Suppose \(A B O C\) is a rhombus in the first quadrant with \(O\) being the origin. If the vertices \(B\) and \(C\) of \(\triangle A B C\) lie respectively on \(y=\frac{4}{3} x\) and \(y=0\) and the side \(B C\) passes through \(\left(\frac{2}{3}, \frac{2}{3}\right)\), then the mid-point of \(B C\) is
- A \(\left(\frac{4}{5}, \frac{2}{5}\right)\)
- B \(\left(\frac{2}{3}, \frac{2}{3}\right)\)
- C \(\left(\frac{2}{5}, \frac{4}{5}\right)\)
- D \(\left(\frac{1}{3}, \frac{1}{3}\right)\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{4}{5}, \frac{2}{5}\right)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \tan 2 \theta=\frac{4}{3}, \tan \theta=x \\ \Rightarrow & \frac{2 x}{1-x^2}=\frac{4}{3} \Rightarrow 2 x^2+3 x-2=0\end{aligned}\) \[ \Rightarrow \quad x=\frac{1}{2} \text { or }-2 \] As \(\theta\) is acute, \(x=\frac{1}{2}=\tan \theta\) \(\therefore\) Slope of…
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