Points X and Y are taken on the sides QR and RS, respectively of a parallelogram PQRS, so that QX=4XR and RY=4YS. The line XY cuts the line PR at Z. Then, PZ is

Options

A. `(21)/(25)PR`
B. `(16)/(25)PR`
C. `(17)/(25)PR`
D. none of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Parallelogram and Points
We have a parallelogram PQRS with points X and Y on sides QR and RS respectively. We know that QX = 4XR and RY = 4YS.

### Step 2: Assign Position Vectors
Let’s assign position vectors:
- Let \( P \) be at the origin, \( P = \vec{0} \).
- Let \( Q \) be at \( \vec{a} \).
- Let \( S \) be at \( \vec{b} \).
- Then, \( R \) will be at \( \vec{a} + \vec{b} \).

### Step 3: Find Position Vector of Point X
Since \( QX = 4XR \), we can use the section formula. The ratio of QX to XR is 4:1, meaning:
\[ \vec{X} = \frac{4 \cdot \vec{R} + 1 \cdot \vec{Q}}{4 + 1} = \frac{4(\vec{a} + \vec{b}) + 1\vec{a}}{5} = \frac{5\vec{a} + 4\vec{b}}{5} \]

### Step 4: Find Position Vector of Point Y
Similarly, for point Y, since \( RY = 4YS \), we have:
\[ \vec{Y} = \frac{4 \cdot \vec{S} + 1 \cdot \vec{R}}{4 + 1} = \frac{4\vec{b} + 1(\vec{a} + \vec{b})}{5} = \frac{\vec{a} + 5\vec{b}}{5} \]

### Step 5: Find Position Vector of Point Z
Let point Z divide line XY in the ratio \( \lambda:1 \). Thus, the position vector of Z can be given by:
\[ \vec{Z} = \frac{\lambda \vec{X} + 1 \cdot \vec{Y}}{\lambda + 1} \]
Substituting the values of \( \vec{X} \) and \( \vec{Y} \):
\[ \vec{Z} = \frac{\lambda \left(\frac{5\vec{a} + 4\vec{b}}{5}\right) + \frac{\vec{a} + 5\vec{b}}{5}}{\lambda + 1} \]
Simplifying this, we get:
\[ \vec{Z} = \frac{\frac{5\lambda \vec{a} + 4\lambda \vec{b} + \vec{a} + 5\vec{b}}{5}}{\lambda + 1} = \frac{(5\lambda + 1)\vec{a} + (4\lambda + 5)\vec{b}}{5(\lambda + 1)} \]

### Step 6: Express Z in terms of P and R
We know that \( \vec{Z} \) can also be expressed as \( \mu \vec{R} \) where \( \vec{R} = \vec{a} + \vec{b} \):
\[ \vec{Z} = \mu(\vec{a} + \vec{b}) \]

### Step 7: Equate the Two Expressions for Z
Equating the two expressions for \( \vec{Z} \):
\[ \frac{(5\lambda + 1)\vec{a} + (4\lambda + 5)\vec{b}}{5(\lambda + 1)} = \mu(\vec{a} + \vec{b}) \]

### Step 8: Compare Coefficients
From the equation above, we can compare coefficients of \( \vec{a} \) and \( \vec{b} \):
1. \( \frac{5\lambda + 1}{5(\lambda + 1)} = \mu \)
2. \( \frac{4\lambda + 5}{5(\lambda + 1)} = \mu \)

Setting these equal gives us:
\[ \frac{5\lambda + 1}{5(\lambda + 1)} = \frac{4\lambda + 5}{5(\lambda + 1)} \]

### Step 9: Solve for λ
Cross-multiplying and simplifying:
\[ 5\lambda + 1 = 4\lambda + 5 \implies \lambda = 4 \]

### Step 10: Find μ
Substituting \( \lambda = 4 \) back into either equation gives:
\[ \mu = \frac{5(4) + 1}{5(4 + 1)} = \frac{21}{25} \]

### Final Step: Conclusion
Thus, the length \( PZ \) in terms of \( PR \) is:
\[ PZ = \frac{21}{25} PR \]