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A, B and C are three points such that \( \overline{AB} = 3a + 4b \) \( \overline{AC} = 15a + 20b \) (a) Prove that A, B and C lie on a straight line - Edexcel - GCSE Maths - Question 16 - 2022 - Paper 1
Question 16
A, B and C are three points such that \( \overline{AB} = 3a + 4b \) \( \overline{AC} = 15a + 20b \) (a) Prove that A, B and C lie on a straight line. D, E and F ... show full transcript
Worked Solution & Example Answer:A, B and C are three points such that \( \overline{AB} = 3a + 4b \) \( \overline{AC} = 15a + 20b \) (a) Prove that A, B and C lie on a straight line - Edexcel - GCSE Maths - Question 16 - 2022 - Paper 1
Step 1
Find the ratio length of DF : length of DE
Answer
Given the conditions for length relationships:
( \overline{DE} = 3e + 6f )
( \overline{EF} = -10.5e - 21f )
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First, express the length of ( DE ): [ |\overline{DE}| = \sqrt{(3e + 6f) \cdot (3e + 6f)} ] This simplifies to a proportional constant; let's denote ( k_1 = 3 ) and we can factor it out.
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Next, express the length of ( EF ): [ |\overline{EF}| = \sqrt{(-10.5e - 21f) \cdot (-10.5e - 21f)} ] The length can also be factored into proportional components.
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The ratio can thus be derived considering the coefficients leading to: [ \frac{|\overline{DF}|}{|\overline{DE}|} = \frac{5}{2} ]
Hence, the ratio is: [ DF : DE = 5:2. ]