To prove that points A, B, and C lie on a straight line, we need to show that the vectors AB and AC are collinear. This means that one vector is a scalar multiple of the other.

Given vectors AB→ = 3a + 4b and AC→ = 15a + 20b, we can see that AC is a scalar multiple of AB.

Divide AC by AB:

(15a + 20b) / (3a + 4b) = 5

Since the ratio is a constant (5), vectors AB and AC are collinear. Therefore, points A, B, and C lie on the same straight line.

Question

To prove that points A, B, and C lie on a straight line, we need to show that the vectors AB and AC are collinear. This means that one vector is a scalar multiple of the other.

Given vectors AB→ = 3a + 4b and AC→ = 15a + 20b, we can see that AC is a scalar multiple of AB.

Divide AC by AB:

(15a + 20b) / (3a + 4b) = 5

Since the ratio is a constant (5), vectors AB and AC are collinear. Therefore, points A, B, and C lie on the same straight line.

Knowee AI · Accepted Answer

To prove that points A, B, and C lie on a straight line, we need to show that the vectors AB and AC are collinear. This means that one vector is a scalar multiple of the other.

Given vectors AB→ = 3a + 4b and AC→ = 15a + 20b, we can see that AC is a scalar multiple of AB.

Divide AC by AB:

(15a + 20b) / (3a + 4b) = 5

Since the ratio is a constant (5), vectors AB and AC are collinear. Therefore, points A, B, and C lie on the same straight line.

A, B and C are three points such thatAB→ = 3a + 4bAC→ = 15a + 20b(a) Prove that A, B and C lie on a straight line

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A, B and C are three points such that

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