NCERT Class X Chapter 7: Coordinate Geometry Exercise 7.1 Question 6(ii)
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: (–3, 5), (3, 1), (0, 3), (–1, –4)
Given:
Points: \( A(-3, 5) \), \( B(3, 1) \), \( C(0, 3) \), \( D(-1, -4) \)
To Find:
The type of quadrilateral formed by the given points.
Formula:
Distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \):
$$ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$Slope of the line joining \( (x_1, y_1) \) and \( (x_2, y_2) \):
$$ m = \frac{y_2 - y_1}{x_2 - x_1} $$Solution:
Step 1: Assign the points as follows:
\( A(-3, 5) \), \( B(3, 1) \), \( C(0, 3) \), \( D(-1, -4) \)
Step 2: Find the lengths of all sides using the distance formula.
Step 3: Calculate \( AB \):
$$ AB = \sqrt{(3 - (-3))^2 + (1 - 5)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} $$Step 4: Calculate \( BC \):
$$ BC = \sqrt{(0 - 3)^2 + (3 - 1)^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} $$Step 5: Calculate \( CD \):
$$ CD = \sqrt{(-1 - 0)^2 + (-4 - 3)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} $$Step 6: Calculate \( DA \):
$$ DA = \sqrt{(-3 - (-1))^2 + (5 - (-4))^2} = \sqrt{(-2)^2 + 9^2} = \sqrt{4 + 81} = \sqrt{85} $$Step 7: Compare the lengths of opposite sides.
\( AB = \sqrt{52} \), \( CD = \sqrt{50} \)
\( BC = \sqrt{13} \), \( DA = \sqrt{85} \)
Opposite sides are not equal.
Step 8: Find the slopes of all sides to check for parallelism.
Step 9: Slope of \( AB \):
$$ m_{AB} = \frac{1 - 5}{3 - (-3)} = \frac{-4}{6} = -\frac{2}{3} $$Step 10: Slope of \( BC \):
$$ m_{BC} = \frac{3 - 1}{0 - 3} = \frac{2}{-3} = -\frac{2}{3} $$Step 11: Slope of \( CD \):
$$ m_{CD} = \frac{-4 - 3}{-1 - 0} = \frac{-7}{-1} = 7 $$Step 12: Slope of \( DA \):
$$ m_{DA} = \frac{5 - (-4)}{-3 - (-1)} = \frac{9}{-2} = -\frac{9}{2} $$Step 13: Check for parallel sides.
\( m_{AB} = m_{BC} = -\frac{2}{3} \), but they are adjacent sides, not opposite.
No pair of opposite sides have equal slopes. So, no sides are parallel.
Step 14: Conclusion.
Since neither the opposite sides are equal nor parallel, the quadrilateral is an irregular quadrilateral.
Result:
The quadrilateral formed by the points \( (-3, 5), (3, 1), (0, 3), (-1, -4) \) is an irregular quadrilateral, as neither the opposite sides are equal nor parallel.
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