MHT CET · Maths · Vector Algebra
\(\mathrm{ABCD}\) is a parallelogram, \(\mathrm{P}\) is the mid-point of \(\mathrm{AB}\). If \(\mathrm{R}\) is the point of intersection
of \(\mathrm{AC}\) and \(\mathrm{DP}\), then \(\mathrm{R}\) divides \(\mathrm{AC}\) internally in the ratio
- A \(3: 1\)
- B \(2: 1\)
- C \(1: 2\)
- D \(2: 3\)
Answer & Solution
Correct Answer
(C) \(1: 2\)
Step-by-step Solution
Detailed explanation
Draw BS parallel to DP as shown
Let \(A P=P B=x \Rightarrow D S=x \Rightarrow S C=x\)
\(\triangle B A Q \sim \triangle P A R\)
\(\therefore \frac{\mathrm{AB}}{\mathrm{AP}}=\frac{\mathrm{AQ}}{\mathrm{AR}} \Rightarrow \frac{2 \mathrm{x}}{\mathrm{x}}=\frac{\mathrm{AQ}}{\mathrm{AR}} \Rightarrow \mathrm{AQ}=2 \mathrm{AR}\)
Thus \(\mathrm{R}\) is mid point of \(\mathrm{AQ}\). i.e. \(\mathrm{AR}=\mathrm{RQ}\)
Similarly \(\Delta\) CQS \(\Delta\) CRD
\(\therefore \mathrm{CQ}=\mathrm{RQ}\)
Thus we get \(\mathrm{AR}=\mathrm{RQ}=\mathrm{CQ}\)
Hence point \(R\) divides \(A C\) in the ratio \(1: 2\)
Let \(A P=P B=x \Rightarrow D S=x \Rightarrow S C=x\)
\(\triangle B A Q \sim \triangle P A R\)
\(\therefore \frac{\mathrm{AB}}{\mathrm{AP}}=\frac{\mathrm{AQ}}{\mathrm{AR}} \Rightarrow \frac{2 \mathrm{x}}{\mathrm{x}}=\frac{\mathrm{AQ}}{\mathrm{AR}} \Rightarrow \mathrm{AQ}=2 \mathrm{AR}\)
Thus \(\mathrm{R}\) is mid point of \(\mathrm{AQ}\). i.e. \(\mathrm{AR}=\mathrm{RQ}\)
Similarly \(\Delta\) CQS \(\Delta\) CRD
\(\therefore \mathrm{CQ}=\mathrm{RQ}\)
Thus we get \(\mathrm{AR}=\mathrm{RQ}=\mathrm{CQ}\)
Hence point \(R\) divides \(A C\) in the ratio \(1: 2\)
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