Question

ABCD is a parallelogram. \(P\) is the midpoint of \(AB\). If \(R\) is the point of intersection of \(AC\) and \(DP\), then \(R\) divides \(AC\) internally in the ratio

• A. \(3:1\)
• B. \(2:1\)
• C. \(1:2\)
• D. \(2:3\)

Hint:

Vector methods are very effective for problems involving ratios of line segments in parallelograms.

Answer 1

The Correct Option is C

Step 1: Use vector representation.
Let the position vectors of points \(A,B,C,D\) be \(\vec{a},\vec{b},\vec{c},\vec{d}\) respectively.
Since \(ABCD\) is a parallelogram, \[ \vec{a}+\vec{c}=\vec{b}+\vec{d} \] Step 2: Find the position vector of point \(P\).
Given \(P\) is the midpoint of \(AB\), \[ \vec{p}=\frac{\vec{a}+\vec{b}}{2} \] Step 3: Parametric form of the lines.
Point \(R\) lies on \(AC\): \[ \vec{r}=\vec{a}+t(\vec{c}-\vec{a}) \] Point \(R\) also lies on \(DP\): \[ \vec{r}=\vec{d}+s(\vec{p}-\vec{d}) \] Step 4: Equate and simplify.
Substitute \(\vec{p}\) and use \(\vec{d}=\vec{a}+\vec{c}-\vec{b}\).
On simplifying, we obtain \[ t=\frac{1}{3} \] Step 5: Determine the ratio.
Thus, \[ AR:RC = t:(1-t) = \frac{1}{3}:\frac{2}{3} = 1:2 \]