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Using section formula and complete the statement. The points $(- 3, - 1), (1, 3)$ and $(- 1, 1)$ are;

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Collinear

Non collinear

Equal

None

answer is A.

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Detailed Solution

Given points are,

A (- 3, - 1), B (1, 3), C (- 1, 1)

Now if a point

(x_{3}, y_{3})

divides the line made by points

(x_{1}, y_{1}) and (x_{2}, y_{2})

in the ratio m:n, then coordinates of

(x_{3}, y_{3})

can be given as,

x_{3} = \frac{m x_{2} + n x_{1}}{m + n}, y_{3} = \frac{m y_{2} + n y_{1}}{m + n}

Suppose that the point C divides AB in ratio

k : 1

. Then
Here

m = k

and

n = 1

so, by the section formula we get

\frac{k - 3}{k + 1} = 1

…..(1) and

\frac{3 k - 1}{k + 1} = 1

….(2)
From equation (2),

\frac{k - 3}{k + 1} = 1

3 k - 1 = k + 1

2 k = 2

k = 1

Thus, the point C divides the line segment AB in the ratio =

1 : 1

Thus, we can say that the points are collinear.
Correct Option is 1.

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