Name the type of triangle formed by the points $ \mathrm{A}(-5,6), \mathrm{B}(-4,-2) $ and $ \mathrm{C}(7,5) $.

Given:

The points \( \mathrm{A}(-5,6), \mathrm{B}(-4,-2) \) and \( \mathrm{C}(7,5) \).

To do:

We have to find the type of triangle formed by the given points.

Solution:

We know that,

The distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

The distance between $A(-5,6)$ and $B(-4,-2) is,

$A B =\sqrt{(-4+5)^{2}+(-2-6)^{2}}$

$=\sqrt{1^{2}+(-8)^{2}}$

$=\sqrt{1+64}$

$=\sqrt{65}$

The distance between $B(-4,-2)$ and $C(7,5)$ is,

$BC =\sqrt{(7+4)^{2}+(5+2)^{2}}$

$=\sqrt{(11)^{2}+(7)^{2}}$

$=\sqrt{121+49}$

$=\sqrt{170}$

The distance between $C(7,5)$ and $A(-5,6)$ is,

$CA=\sqrt{(-5-7)^{2}+(6-5)^{2}}$

$=\sqrt{(-12)^{2}+1^{2}}$

$=\sqrt{144+1}$

$=\sqrt{145}$

$AB^2+CA^2=(\sqrt{65})^2+(\sqrt{145})^2$

$=65+145$

$=210$

$BC^2=(\sqrt{170})^2$

$=170$

Here,

$AB ≠ BC ≠ CA$ and $AB^2+CA^2≠BC^2$

Therefore, the points \( \mathrm{A}(-5,6), \mathrm{B}(-4,-2) \) and \( \mathrm{C}(7,5) \) form a scalene triangle. 

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Updated on: 10-Oct-2022

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