Prove that the points $(2, 3), (-4, -6)$ and $(1, \frac{3}{2} )$ do not form a triangle.

Given:

Given points are $(2, 3), (-4, -6)$ and $(1, \frac{3}{2} )$.

To do:

We have to prove that the given points do not form a triangle.

Solution:

Let the three points be \( \mathrm{A}(2,3), \mathrm{B}(-4,-6) \) and \( C(1,\frac{3}{2}) \).

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,

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

\( =\sqrt{(-4-2)^{2}+(-6-3)^{2}} \)

\( =\sqrt{(-6)^{2}+(-9)^{2}} \)

\( =\sqrt{36+81}=\sqrt{117} \)

\( \mathrm{BC}=\sqrt{(1+4)^{2}+(\frac{3}{2}+6)^{2}} \)

\( =\sqrt{(5)^{2}+(\frac{3+12}{2})^{2}} \)

\( =\sqrt{25+\frac{225}{4}} \)

\( =\sqrt{\frac{25(4)+225}{4}} \)

\( =\sqrt{\frac{100+225}{4}} \)

\( =\sqrt{\frac{325}{4}} \)

\( \mathrm{CA}=\sqrt{(2-1)^{2}+(3-\frac{3}{2})^{2}} \)

\( =\sqrt{(1)^{2}+(\frac{3(2)-3}{2})^{2}} \)

\( =\sqrt{1+\frac{9}{4}} \)

\( =\sqrt{\frac{4(1)+9}{4}} \)

\( =\sqrt{\frac{13}{4}} \)

Here,

\( \mathrm{BC}+\mathrm{CA}=\sqrt{\frac{325}{4}}+\sqrt{\frac{13}{4}} \)

\( \mathrm{AB}=\sqrt{117} \)

But, \(  \sqrt{\frac{325}{4}}+\sqrt{\frac{13}{4}}<\sqrt{117} \)

Therefore, the given points do not form a triangle.

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Updated on: 10-Oct-2022

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