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A, B and C are three points such that AB → = 3a + 4b AC → = 15a + 20b (a) Prove that A, B and C lie on a straight line - Edexcel - GCSE Maths - Question 15 - 2022 - Paper 1
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A, B and C are three points such that AB → = 3a + 4b AC → = 15a + 20b (a) Prove that A, B and C lie on a straight line. D, E and F are three points on a straight... show full transcript
Worked Solution & Example Answer:A, B and C are three points such that AB → = 3a + 4b AC → = 15a + 20b (a) Prove that A, B and C lie on a straight line - Edexcel - GCSE Maths - Question 15 - 2022 - Paper 1
Step 1
(a) Prove that A, B and C lie on a straight line.
Answer
To show that points A, B, and C are collinear, we need to demonstrate that the vectors AB and AC are scalar multiples of each other.
Starting with the vectors:
- AB → = 3a + 4b
- AC → = 15a + 20b
We can express AC → in terms of AB →. If we factor out 5 from AC, we have:
Since AC is a scalar multiple of AB, this confirms that A, B, and C lie on a straight line.
Step 2
(b) Find the ratio length of DF : length of DE
Answer
To find the ratio of the lengths of DF and DE, we first express the vectors DE and EF:
- DE → = 3e + 6f
- EF → = -10.5e - 21f
The vector DF can be calculated as:
Substituting the vectors:
DF → = (3 - 10.5)e + (6 - 21)f \ DF → = -7.5e - 15f $$ Next, we need to calculate the lengths: - Length of DE: $$ |DE| = \sqrt{(3)^2 + (6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} $$ - Length of DF: $$ |DF| = \sqrt{(-7.5)^2 + (-15)^2} = \sqrt{56.25 + 225} = \sqrt{281.25} = 7.5\sqrt{5} $$ Now we find the ratio: $$ \frac{|DF|}{|DE|} = \frac{7.5\sqrt{5}}{3\sqrt{5}} = \frac{7.5}{3} = 2.5 $$ Thus, the ratio of lengths DF to DE is: $$ DF:DE = 2.5:1 = 5:2 $$