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

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.

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Updated on: 10-Oct-2022

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