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OABC is a parallelogram - Edexcel - GCSE Maths - Question 19 - 2017 - Paper 1

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OABC is a parallelogram. $$\vec{OA} = \vec{a}$$ and $$\vec{OC} = \vec{c}$$ X is the midpoint of the line AC. OC is a straight line so that OC : CD = k : 1. Given tha... show full transcript

Worked Solution & Example Answer:OABC is a parallelogram - Edexcel - GCSE Maths - Question 19 - 2017 - Paper 1

Step 1

X is the midpoint of the line AC.

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Answer

Since X is the midpoint of AC, we have: X⃗=A⃗+C⃗2=a⃗+c⃗2\vec{X} = \frac{\vec{A} + \vec{C}}{2} = \frac{\vec{a} + \vec{c}}{2}

Step 2

OC is a straight line so that OC : CD = k : 1.

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Answer

Let ( \vec{D} ) be the point on line OC. From the ratio OC : CD = k : 1, we can express: D⃗=C⃗+1k(C⃗−O⃗)=c⃗+1k(c⃗−0⃗)=c⃗(1+1k)\vec{D} = \vec{C} + \frac{1}{k} (\vec{C} - \vec{O}) = \vec{c} + \frac{1}{k} (\vec{c} - \vec{0}) = \vec{c}(1 + \frac{1}{k})

Step 3

Given that \(\vec{XD} = 3\vec{e} - \frac{1}{2}\vec{a}\).

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We can express (\vec{XD}) as: XD⃗=D⃗−X⃗\vec{XD} = \vec{D} - \vec{X} Substituting the expressions for (\vec{D}) and (\vec{X}): XD⃗=c⃗(1+1k)−a⃗+c⃗2\vec{XD} = \vec{c}(1 + \frac{1}{k}) - \frac{\vec{a} + \vec{c}}{2} Setting this equal to the given expression, we have: c⃗(1+1k)−a⃗+c⃗2=3e⃗−12a⃗\vec{c}(1 + \frac{1}{k}) - \frac{\vec{a} + \vec{c}}{2} = 3\vec{e} - \frac{1}{2}\vec{a}

Step 4

Find the value of k.

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Answer

By equating the coefficients, we derive: For the vector parts, isolating k leads to: k=2.5k = 2.5 Thus, the value of k is 2.5.

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