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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.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B.
Solution :
Given: Two points in a Cartesian plane, $P(x_1, y_1) = (0, 0)$ and $Q(x_2, y_2) = (36, 15)$.
To Find: The distance between points $P$ and $Q$, and subsequently, the distance between two towns $A$ and $B$ represented by these coordinates.
Step 1: State the Distance Formula
The distance $d$ between any two points $(x_1, y_1)$ and $(x_2, y_2)$ in a coordinate plane 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 right-angled triangle formed by the horizontal and vertical differences of the coordinates.]
Step 2: Substitute the Given Values
Assign the coordinates:
$x_1 = 0, y_1 = 0$
$x_2 = 36, y_2 = 15$
Substituting these into the formula:
$d = \sqrt{(36 - 0)^2 + (15 - 0)^2}$
Step 3: Perform Algebraic Simplification
Calculate the squares of the differences:
$d = \sqrt{(36)^2 + (15)^2}$
Calculate $36^2$: $36 \times 36 = 1296$
Calculate $15^2$: $15 \times 15 = 225$
Sum the squares:
$d = \sqrt{1296 + 225}$
$d = \sqrt{1521}$
Step 4: Extract the Square Root
To find $\sqrt{1521}$, we look for a number which, when multiplied by itself, equals $1521$.
Since $30^2 = 900$ and $40^2 = 1600$, the value must be between 30 and 40.
Testing $39$: $39 \times 39 = 1521$.
$d = 39$
Step 5: Application to Towns A and B
If town $A$ is located at $(0, 0)$ and town $B$ is located at $(36, 15)$, the distance between them is equivalent to the distance calculated between points $P$ and $Q$.
Therefore, the distance between town $A$ and town $B$ is $39$ units.
Final Answer: The distance between the points (0, 0) and (36, 15) is 39 units. Consequently, the distance between town A and town B is 39 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)
- 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).
- 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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