Three points A (1, -2), B(3, 4) and C(4, 7) form:

a straight line

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an equilateral triangle

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a right angled triangle

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none of (a), (b), (c)

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Solution

The correct option is A a straight line
$m_{1} = S l o p e o f A B = \frac{4 - (- 2)}{3 - 1} = 3 a n d m_{2} = S l o p e o f B C = \frac{7 - 4}{4 - 3} = 3 ∴ m_{1} = m_{2}$
$∴$ AB is parallel to BC and B is common to both AB & BC.
Hence, the point A(1, -2), B(3, 4) and C(4, 7) are collinear.

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