Using slope concept show that the points A(-2, -2), B(1, 1) and D(3, 3)

are not collinear

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cannot be plotted

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are collinear

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are the same point on the graph

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Solution

The correct option is C are collinear
$T h e g i v e n p o i n t s a r e A (x_{1}, y_{1}) = (2, - 2), B (x_{2}, y_{2}) = (1, 1) & D (x_{3}, y_{3}) = (3, 3) . N o w i f 3 p o i n t s A (x_{1}, y_{1}), B (x_{2}, y_{2}) & D (x_{3}, y_{3}) a r e c o l l i n e a r t h e n s l o p e m_{A B} = s l o p e m_{B D} = s l o p e m_{A D} . . s l o p e m_{A B} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{1 + 2}{1 + 2} = 1, s l o p e m_{B D} = \frac{y_{3} - y_{2}}{x_{3} - x_{2}} = \frac{3 - 1}{3 - 1} = 1 a n d s l o p e m_{A D} = \frac{y_{3} - y_{1}}{x_{3} - x_{1}} = \frac{3 + 2}{3 + 2} = 1. ∴ s l o p e m_{A B} = s l o p e m_{B D} = s l o p e m_{A D} . ∴ T h e p o i n t s A, B & D a r e c o l l i n e a r . A n s - O p t i o n C .$

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A (5, - 2), B (4, - 1)

C (1, 2)

(- 2, - 1), (4, 0), (3, 3)

(- 3, 2)

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