Prove that the point $(-2, 5), (0, 1)$ and $(2, -3)$ are collinear.

Given:

Given points are $(-2, 5), (0, 1)$ and $(2, -3)$.

To do:

We have to prove that the points $(-2, 5), (0, 1)$ and $(2, -3)$ are collinear.

Solution:

Let the points be \( \mathrm{A}(-2,5), \mathrm{B}(0,1) \) and \( \mathrm{C} (2,-3) \)

We know that,

The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).

Therefore,

\( \mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \)
\( =\sqrt{(0+2)^{2}+(1-5)^{2}} \)
\( =\sqrt{(2)^{2}+(-4)^{2}} \)

\( =\sqrt{4+16} \)
\( =\sqrt{20} \)

\( =\sqrt{4 \times 5} \)

\( =2 \sqrt{5} \)
\( \mathrm{BC}=\sqrt{(2-0)^{2}+(-3-1)^{2}} \)
\( =\sqrt{(2)^{2}+(-4)^{2}} \)
\( =\sqrt{4+16} \)

\( =\sqrt{20} \)

\( =\sqrt{4 \times 5} \)

\( =2 \sqrt{5} \)
\( \mathrm{CA}=\sqrt{(-2-2)^{2}+(5+3)^{2}} \)
\( =\sqrt{(-4)^{2}+(8)^{2}} \)

\( =\sqrt{16+64} \)
\( =\sqrt{80} \)

\( =\sqrt{16 \times 5} \)

\( =4 \sqrt{5} \)

Here,

$AB + BC = 2\sqrt5 +2\sqrt5$
$CA = 4\sqrt5$
$\Rightarrow AB + BC = CA$
Therefore, the points $A, B$ and $C$ are collinear.

Hence proved.

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Updated on: 10-Oct-2022

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