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Question
Using the method of slope, show that the following points are collinear A (16, − 18), B (3, −6), C (−10, 6) .
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Solution
A (16, − 18), B (3, −6), C (−10, 6)
Slope of AB = \[\frac{y_2 - y_1}{x_2 - x_1} = \frac{- 6 + 18}{3 - 16} = - \frac{12}{13}\]
Slope of BC = \[\frac{y_2 - y_1}{x_2 - x_1} = \frac{6 + 6}{- 10 - 3} = - \frac{12}{13}\]
Since, Slope of AB = Slope of BC = \[- \frac{12}{13}\]
Therefore, the given points are collinear.
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| Column C1 | Column C2 |
| (a) The coordinates of the points P and Q on the line x + 5y = 13 which are at a distance of 2 units from the line 12x – 5y + 26 = 0 are |
(i) (3, 1), (–7, 11) |
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(ii) `(- 1/3, 11/3), (4/3, 7/3)` |
| (c) The coordinates of the point on the line joining A (–2, 5) and B (3, 1) such that AP = PQ = QB are |
(iii) `(1, 12/5), (-3, 16/5)` |
