TS EAMCET · Maths · Three Dimensional Geometry
\(A B C D\) is a parallelogram and \(P\) is a point on the segment \(\mathbf{A D}\) dividing it internally in the ratio \(3: 1\). If the line \(\mathbf{B P}\) meets the diagonal \(A C\) in \(Q\), then \(A Q: Q C\) equals
- A \(3: 4\)
- B \(4: 3\)
- C \(3: 2\)
- D \(2: 3\)
Answer & Solution
Correct Answer
(A) \(3: 4\)
Step-by-step Solution
Detailed explanation
Since, \(P\) divides \(A D\) in the ratio \(3: 1\), so \(P\) will be \(\frac{3 d}{4}\) Let \(Q\) divides \(B P\) in \(1: \lambda\) and \(A C\) in \(1: \mu\).…
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