NCERT Class X Chapter 7: Coordinate Geometry Exercise 7.1 Question 6(i)

NCERT Class X Chapter 7: Coordinate Geometry

Question:

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (–1, –2), (1, 0), (–1, 2), (–3, 0)

Given:

Points: \( A(-1, -2) \), \( B(1, 0) \), \( C(-1, 2) \), \( D(-3, 0) \)

To Find:

The type of quadrilateral formed by the given points.

Formula:

Distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \):

$$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Solution:

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

$$ AB = \sqrt{(1 - (-1))^2 + (0 - (-2))^2} = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} $$

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

$$ BC = \sqrt{((-1) - 1)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} $$

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

$$ CD = \sqrt{((-3) - (-1))^2 + (0 - 2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} $$

Step 4: Find the length of side \( DA \).

$$ DA = \sqrt{((-1) - (-3))^2 + ((-2) - 0)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} $$

Step 5: All four sides are equal. Therefore, the quadrilateral is either a rhombus or a square. Let's check the diagonals.

Step 6: Find diagonal \( AC \).

$$ AC = \sqrt{((-1) - (-1))^2 + (2 - (-2))^2} = \sqrt{0^2 + 4^2} = \sqrt{16} = 4 $$

Step 7: Find diagonal \( BD \).

$$ BD = \sqrt{((-3) - 1)^2 + (0 - 0)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4 $$

Step 8: Both diagonals are equal, and all sides are equal. Therefore, the quadrilateral is a square.

Result:

The quadrilateral formed by the given points is a square.

© Kaliyuga Ekalavya. All rights reserved.

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