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Q3:
Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.
Solution :
Given: Three points in a Cartesian plane: $A(1, 5)$, $B(2, 3)$, and $C(-2, -11)$.
To Find: Determine whether the points $A$, $B$, and $C$ are collinear.
Conceptual Framework: Three points are said to be collinear if they lie on the same straight line. For three points $A$, $B$, and $C$, they are collinear if and only if the sum of the lengths of two smaller segments equals the length of the longest segment (e.g., $AB + BC = AC$, or any permutation thereof).
Formula Used: The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the Distance Formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Step 1: Calculate the distance between points $A(1, 5)$ and $B(2, 3)$
Let $(x_1, y_1) = (1, 5)$ and $(x_2, y_2) = (2, 3)$.
$AB = \sqrt{(2 - 1)^2 + (3 - 5)^2}$
$AB = \sqrt{(1)^2 + (-2)^2}$
$AB = \sqrt{1 + 4}$
$AB = \sqrt{5}$ units
Step 2: Calculate the distance between points $B(2, 3)$ and $C(-2, -11)$
Let $(x_1, y_1) = (2, 3)$ and $(x_2, y_2) = (-2, -11)$.
$BC = \sqrt{(-2 - 2)^2 + (-11 - 3)^2}$
$BC = \sqrt{(-4)^2 + (-14)^2}$
$BC = \sqrt{16 + 196}$
$BC = \sqrt{212}$
$BC = \sqrt{4 \times 53} = 2\sqrt{53}$ units
Step 3: Calculate the distance between points $A(1, 5)$ and $C(-2, -11)$
Let $(x_1, y_1) = (1, 5)$ and $(x_2, y_2) = (-2, -11)$.
$AC = \sqrt{(-2 - 1)^2 + (-11 - 5)^2}$
$AC = \sqrt{(-3)^2 + (-16)^2}$
$AC = \sqrt{9 + 256}$
$AC = \sqrt{265}$ units
Step 4: Verify the Collinearity Condition
We check if the sum of any two distances equals the third:
$AB + BC = \sqrt{5} + 2\sqrt{53} \approx 2.236 + 14.56 = 16.796$
$AC = \sqrt{265} \approx 16.279$
Since $AB + BC \neq AC$, $AB + AC \neq BC$, and $BC + AC \neq AB$, the condition for collinearity is not satisfied.
Final Answer: The points $(1, 5)$, $(2, 3)$, and $(-2, -11)$ are not collinear.
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.
- 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)
- 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)
- 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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