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Question
Without using distance formula, prove that the points A(1, 4), B(3, –2) and C(–3, 16) are collinear.
Theorem
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Solution
Given: Points A(1, 4), B(3, –2) and C(–3, 16).
To prove: A, B, C are collinear.
Proof:
1. Three points A, B, C are collinear if and only if slope of AB = slope of BC. [Condition for collinearity]
2. Slope of AB:
$$m_{AB} = \frac{-2 - 4}{3 - 1}$$
$$= \frac{-6}{2}$$
$$= -3$$
3. Slope of BC:
$$m_{BC} = \frac{16 - (-2)}{-3 - 3}$$
$$= \frac{18}{-6}$$
$$= -3$$
4. Since slope of AB = slope of BC = –3, AB is parallel to BC. Since both lines pass through B, they coincide. Hence A, B, C are collinear.
Hence proved.
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