NCERT Class X Chapter 7: Coordinate Geometry Example 2
NCERT Class X Chapter 7: Coordinate Geometry Example 2
Question:
Show that the points (1, 7), (4, 2), (–1, –1) and (–4, 4) are the vertices of a square.
Given:
The points are A(1, 7), B(4, 2), C(–1, –1), and D(–4, 4).
To Find:
Show that these points are the vertices of a square.
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 AB.
$$ AB = \sqrt{(4 - 1)^2 + (2 - 7)^2} = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34} $$Step 2: Find the length of BC.
$$ BC = \sqrt{(-1 - 4)^2 + (-1 - 2)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} $$Step 3: Find the length of CD.
$$ CD = \sqrt{(-4 - (-1))^2 + (4 - (-1))^2} = \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} $$Step 4: Find the length of DA.
$$ DA = \sqrt{(1 - (-4))^2 + (7 - 4)^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} $$Step 5: Find the length of diagonal AC.
$$ AC = \sqrt{(-1 - 1)^2 + (-1 - 7)^2} = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68} $$Step 6: Find the length of diagonal BD.
$$ BD = \sqrt{(-4 - 4)^2 + (4 - 2)^2} = \sqrt{(-8)^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68} $$Step 7: Check the properties of a square.
All four sides are equal: \( AB = BC = CD = DA = \sqrt{34} \).
Both diagonals are equal: \( AC = BD = \sqrt{68} \).
Also, \( (\sqrt{68})^2 = 68 = 2 \times 34 = 2 \times (\sqrt{34})^2 \).
Step 8: Conclusion.
Since all sides are equal and both diagonals are equal and longer than the sides, the points are vertices of a square.
Result:
The points (1, 7), (4, 2), (–1, –1) and (–4, 4) are the vertices of a square.
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