# The points $(-4,0),(4,0),(0,3)$ are the vertices of a(A) right triangle(B) isosceles triangle(C) equilateral triangle(D) scalene triangle

#### Complete Python Prime Pack

9 Courses     2 eBooks

#### Artificial Intelligence & Machine Learning Prime Pack

6 Courses     1 eBooks

#### Java Prime Pack

9 Courses     2 eBooks

Given:

Given points are $(-4,0),(4,0),(0,3)$

To do:

We have to choose the correct option.

Solution:

Let the vertices of the $\Delta \mathrm{ABC}$ be $\mathrm{A}(-4,0), \mathrm{B}(4,0)$ and $\mathrm{C}(0,3)$.

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,

$A B=\sqrt{[4-(-4)]^{2}+(0-0)^{2}}$

$=\sqrt{(4+4)^{2}+(0)^{2}}$

$=\sqrt{8^2}$

$=8$

$B C=\sqrt{(0-4)^{2}+(3-0)^{2}}$

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

$=\sqrt{16+9}$

$=\sqrt{25}$

$=5$

$CA=\sqrt{(-4-0)^{2}+(0-3)^{2}}$

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

$=\sqrt{16+9}$

$=\sqrt{25}$

$=5$

Here,

$\mathrm{BC}=\mathrm{CA}$ and $\mathrm{AB}$ is the longest side.

$\mathrm{BC}^{2}+\mathrm{CA}^{2}=(5)^{2}+(5)^{2}$

$=25+25=50$

$\mathrm{AB}^{2}=(8)^{2}=64$

$\therefore \mathrm{BC}^{2}+\mathrm{CA}^{2}≠\mathrm{AB}^{2}$

Therefore, $\Delta \mathrm{ABC}$ is an isosceles triangle.

Updated on 10-Oct-2022 13:28:26