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Q1(iii):
Find the distance between the following pairs of points : (iii) ($a$, $b$), (– $a$, – $b$)
Solution :
Given: Two points in a Cartesian plane, $P(x_1, y_1) = (a, b)$ and $Q(x_2, y_2) = (-a, -b)$.
To find: The distance between the points $P$ and $Q$.
Visual Representation:
Step 1: State the Distance Formula
The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the Euclidean distance formula:
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
[This formula is derived from the Pythagorean Theorem applied to the horizontal and vertical differences between coordinates.]
Step 2: Substitute the given coordinates into the formula
Given $x_1 = a$, $y_1 = b$, $x_2 = -a$, and $y_2 = -b$.
Substituting these values into the distance formula:
$d = \sqrt{(-a - a)^2 + (-b - b)^2}$
Step 3: Simplify the expressions inside the parentheses
Perform the subtraction within the brackets:
$(-a - a) = -2a$
$(-b - b) = -2b$
Therefore:
$d = \sqrt{(-2a)^2 + (-2b)^2}$
Step 4: Evaluate the squares
Recall that $(-x)^2 = x^2$:
$(-2a)^2 = 4a^2$
$(-2b)^2 = 4b^2$
Substituting these back into the equation:
$d = \sqrt{4a^2 + 4b^2}$
Step 5: Factor out the common term and simplify the radical
Factor out $4$ from the terms inside the square root:
$d = \sqrt{4(a^2 + b^2)}$
Using the property of radicals $\sqrt{xy} = \sqrt{x} \cdot \sqrt{y}$:
$d = \sqrt{4} \cdot \sqrt{a^2 + b^2}$
Since $\sqrt{4} = 2$:
$d = 2\sqrt{a^2 + b^2}$
Final Answer: The distance between the points $(a, b)$ and $(-a, -b)$ is $2\sqrt{a^2 + b^2}$ units.
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)
- 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)
- 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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