NCERT Class X Chapter 7: Coordinate Geometry Exercise 7.1 Question 6(iii)
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: (4, 5), (7, 6), (4, 3), (1, 2)
Given:
Points: \( A(4, 5),\ B(7, 6),\ C(4, 3),\ D(1, 2) \)
To Find:
The type of quadrilateral formed by the given points.
Formula:
Distance formula:
\[
\text{Distance between } (x_1, y_1) \text{ and } (x_2, y_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Solution:
Step 1: Assign the points as \( A(4,5) \), \( B(7,6) \), \( C(4,3) \), \( D(1,2) \).
Step 2: Find the length of \( AB \):
\[ AB = \sqrt{(7-4)^2 + (6-5)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \]Step 3: Find the length of \( BC \):
\[ BC = \sqrt{(4-7)^2 + (3-6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} \]Step 4: Find the length of \( CD \):
\[ CD = \sqrt{(1-4)^2 + (2-3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \]Step 5: Find the length of \( DA \):
\[ DA = \sqrt{(4-1)^2 + (5-2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} \]Step 6: Compare the lengths of opposite sides:
\[ AB = CD = \sqrt{10}, \quad BC = DA = \sqrt{18} \]Both pairs of opposite sides are equal.
Step 7: Find the lengths of the diagonals \( AC \) and \( BD \):
\[ AC = \sqrt{(4-4)^2 + (3-5)^2} = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2 \] \[ BD = \sqrt{(1-7)^2 + (2-6)^2} = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} \]Step 8: Since both pairs of opposite sides are equal, but diagonals are not equal, the quadrilateral is a parallelogram (not a rectangle or square).
Result:
The quadrilateral formed by the given points is a parallelogram.
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