OYZ is a parallelogram - Edexcel - GCSE Maths - Question 1 - 2019 - Paper 3

For point R, we analyze the ratio ( OR : RY = 1 : 3 ). Thus, we have:

$\overrightarrow{OR} = \frac{1}{4} \overrightarrow{OY} = \frac{1}{4} \mathbf{b}$

Similarly, the vector for segment ( \overrightarrow{RY} ) is:

$\overrightarrow{RY} = \overrightarrow{OY} - \overrightarrow{OR} = \mathbf{b} - \frac{1}{4} \mathbf{b} = \frac{3}{4} \mathbf{b}$

Next, we express the vectors from Z to P and Z to R:

1. The vector ( \overrightarrow{ZP} ): $\overrightarrow{ZP} = \overrightarrow{ZY} - \overrightarrow{YP} = (\overrightarrow{OY} + \overrightarrow{XZ}) - \overrightarrow{OP} = \mathbf{b} + (\mathbf{a} - 0) - \frac{1}{3} \mathbf{a} = \mathbf{b} + \frac{2}{3} \mathbf{a}$

2. The vector ( \overrightarrow{ZR} ): $\overrightarrow{ZR} = \overrightarrow{ZY} - \overrightarrow{YR} = (\overrightarrow{OY} + \overrightarrow{XZ}) - \overrightarrow{OR} = \mathbf{b} + (\mathbf{a} - 0) - \frac{1}{4} \mathbf{b} = \mathbf{b} + \frac{3}{4} \mathbf{b} = \frac{7}{4} \mathbf{b}$

We now have both vectors:

• ( ZP = \mathbf{b} + \frac{2}{3} \mathbf{a} )
• ( ZR = \frac{7}{4} \mathbf{b} )

To find the ratio ( ZP : ZR ), we write: $\frac{ZP}{ZR} = \frac{\left( \mathbf{b} + \frac{2}{3} \mathbf{a} \right)}{\frac{7}{4} \mathbf{b}} \text{, multiplying both parts by } 4 \text{ to simplify the fractions.}$

Simplifying gives: $ZP : ZR = 4\left( \frac{4}{7}\right)\mathbf{b} : \mathbf{b} + \frac{2}{3}\mathbf{a}$