NCERT Class X Chapter 7: Coordinate Geometry Exercise 7.1 Question 4

NCERT Class X Chapter 7: Coordinate Geometry

Question:

Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.

Given:

Points: \( A(5, -2) \), \( B(6, 4) \), \( C(7, -2) \).

To Find:

Whether triangle \( ABC \) is an isosceles triangle.

Formula:

Distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \):
$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Solution:

Step 1: Find the length of side \( AB \).

$$ AB = \sqrt{(6 - 5)^2 + [4 - (-2)]^2} = \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37} $$

Step 2: Find the length of side \( BC \).

$$ BC = \sqrt{(7 - 6)^2 + [(-2) - 4]^2} = \sqrt{1^2 + (-6)^2} = \sqrt{1 + 36} = \sqrt{37} $$

Step 3: Find the length of side \( CA \).

$$ CA = \sqrt{(5 - 7)^2 + [(-2) - (-2)]^2} = \sqrt{(-2)^2 + 0^2} = \sqrt{4 + 0} = 2 $$

Step 4: Compare the side lengths.

We have \( AB = \sqrt{37} \), \( BC = \sqrt{37} \), and \( CA = 2 \).

Since two sides are equal, the triangle is isosceles.

Result:

Therefore, the triangle with vertices \( (5, -2) \), \( (6, 4) \), and \( (7, -2) \) is an isosceles triangle.

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