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Question
Determine whether the following point is collinear.
D(–2, –3), E(1, 0), F(2, 1)
Solution
D(–2, –3), E(1, 0), F(2, 1)
\[\mathrm{Slope~of~line~DE}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} =\frac{0-\left(-3\right)}{1-\left(-2\right)}\]
= \[\frac{0 + 3}{1 + 2} = \frac{3}{3} = 1\]
\[\mathrm{Slope~of~line~EF}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{1-0}{2-1}=1\]
∴ Slope of DE = Slope of EF = 1
∴ line DE || line EF
Also, point E is common to both the lines.
∴ Both lines are the same.
∴ Points D, E and F are collinear.
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Solution:
Slope of line = `("y"_2 - "y"_1)/("x"_2 - "x"_1)`
∴ Slope of line AB = `(2 - 1)/(8 - 6) = square` .......(i)
∴ Slope of line BC = `(4 - 2)/(9 - 8) = square` .....(ii)
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∴ Slope of line AB = `square` ......[From (i) and (iii)]
∴ line AB || line CD
∴ Slope of line BC = `square` ......[From (ii) and (iv)]
∴ line BC || line DA
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∴ `square`ABCD is a parallelogram.
Four alternative answers for the following question are given. Choose the correct alternative and write its alphabet:
Slope of X-axis is ______.
