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Q6(iii):
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution :
Given: The coordinates of the four points are $A(4, 5)$, $B(7, 6)$, $C(4, 3)$, and $D(1, 2)$.
To Find: The type of quadrilateral formed by these points, if any, and the reasons for the classification.
Step 1: Formula Definition
To determine the nature of the quadrilateral, we use the Distance Formula to find the lengths of the four sides ($AB, BC, CD, DA$) and the two diagonals ($AC, BD$).
The Distance Formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Step 2: Calculating Side Lengths
- Length of AB: $\sqrt{(7 - 4)^2 + (6 - 5)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$ units.
- Length of BC: $\sqrt{(4 - 7)^2 + (3 - 6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.
- Length of CD: $\sqrt{(1 - 4)^2 + (2 - 3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$ units.
- Length of DA: $\sqrt{(4 - 1)^2 + (5 - 2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.
Step 3: Calculating Diagonal Lengths
- Length of AC: $\sqrt{(4 - 4)^2 + (3 - 5)^2} = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2$ units.
- Length of BD: $\sqrt{(1 - 7)^2 + (2 - 6)^2} = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$ units.
Step 4: Analysis and Conclusion
From the calculations:
1. Opposite sides are equal: $AB = CD = \sqrt{10}$ and $BC = DA = 3\sqrt{2}$.
2. Diagonals are unequal: $AC = 2$ and $BD = \sqrt{52}$.
Since the opposite sides are equal and the diagonals are not equal, the quadrilateral satisfies the properties of a parallelogram.
Final Answer: The points form a parallelogram.
More Questions from Class 10 Mathematics Coordinate geometry EXERCISE 7.1
- Q1(i): Find the distance between the following pairs of points : (i) (2, 3), (4, 1)
- Q1(ii): Find the distance between the following pairs of points : (ii) (– 5, 7), (– 1, 3)
- Q1(iii): Find the distance between the following pairs of points : (iii) ($a$, $b$), (– $a$, – $b$)
- Q10: Find a relation between $x$ and $y$ such that the point ($x$, $y$) is equidistant from the point (3, 6) and (– 3, 4).
- Q2: Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B.
- Q3: Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.
- Q4: Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
- Q5: In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
- Q6(i): Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)
- Q6(ii): Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (ii) (–3, 5), (3, 1), (0, 3), (–1, – 4)
- Q7: Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
- Q8: Find the values of $y$ for which the distance between the points P(2, – 3) and Q(10, $y$) is 10 units.
- Q9: If Q(0, 1) is equidistant from P(5, –3) and R($x$, 6), find the values of $x$. Also find the distances QR and PR.
CBSE Solutions for Class 10 Mathematics Coordinate geometry
Chapters in CBSE - Class 10 Mathematics
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