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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

Solution:

Consider A(-4, 0), B(4, 0), C(0, 3) as the vertices given

The points (–4, 0), (4, 0), (0, 3) are the vertices of a

We know that the formula to find the distance between two points (x₁, y₁) and (x₂, y₂) is

\(Distance=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\)

Now distance between A and B

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

AB = √64 = 8

Distance between B and C

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

BC = √25 = 5

Distance between A and C

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

AC =√25 = 5

As BC = AC the triangle is isosceles

Therefore, the points are the vertices of an isosceles triangle.

✦ Try This: The points (-2, 0), (2, 0), (0, 1) are the vertices of a

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7

NCERT Exemplar Class 10 Maths Exercise 7.1 Problem 8

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

Summary:

The points (–4, 0), (4, 0), (0, 3) are the vertices of an isosceles triangle

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