Coordinate Geometry | Exercise 7.1

Question 2

Find the distance between the points (0,0)(0, 0) and (36,15)(36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution

Distance Formula: The distance between two points A(x1,y1)\text{A}(x_1, y_1) and B(x2,y2)\text{B}(x_2, y_2) is given by:

d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

This formula is derived from the Pythagoras theorem, where the distance is the hypotenuse of a right triangle formed by the horizontal and vertical differences between the two points.

We will use the distance formula to find the distance between two points.

Step 1 — Find distance between points

Let's label the given points. Point P is (0,0)(\mathbf{0}, \mathbf{0}). Point Q is (36,15)(\mathbf{36}, \mathbf{15}). We use the distance formula. Distance PQ=((x2−x1)2+(y2−y1)2)PQ = \sqrt{((x_2 - x_1)^2 + (y_2 - y_1)^2)}.

PQ=((36−0)2+(15−0)2)PQ = \sqrt{((36 - 0)^2 + (15 - 0)^2)}

=(362+152)= \sqrt{(36^2 + 15^2)}

=(1296+225)= \sqrt{(1296 + 225)}

=1521= \sqrt{1521}

39 units\boxed{39 \text{ units}}

Diagram 1

Step 2 — Distance between towns

Yes, we can find the distance. The distance between towns A and B is the same. We use the same distance formula. The distance is 39 units.

Answer

(i) The distance between (0,0)(\mathbf{0}, \mathbf{0}) and (36,15)(\mathbf{36}, \mathbf{15}) is 39\mathbf{39} units. (ii) Yes, the distance between towns A and B is 39\mathbf{39} units.

More questions in Exercise 7.1

Q1

Find the distance between the following pairs of points :

(i) (2,3),(4,1)(2, 3), (4, 1) (ii) (−5,7),(−1,3)(-5, 7), (-1, 3) (iii) (a,b),(−a,−b)(a, b), (-a, -b)

Q2

Find the distance between the points (0,0)(0, 0) and (36,15)(36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

Q3

Determine if the points (1,5),(2,3)(1, 5), (2, 3) and (−2,−11)(-2, -11) are collinear.

Q4

Check whether (5,−2),(6,4)(5, -2), (6, 4) and (7,−2)(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

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)(-1, -2), (1, 0), (-1, 2), (-3, 0)

(ii) (−3,5),(3,1),(0,3),(−1,−4)(-3, 5), (3, 1), (0, 3), (-1, -4)

(iii) (4,5),(7,6),(4,3),(1,2)(4, 5), (7, 6), (4, 3), (1, 2)

Q7

Find the point on the xx-axis which is equidistant from (2,−5)(2, -5) and (−2,9)(-2, 9).

Q8

Find the values of yy for which the distance between the points P(2,−3)P(2, -3) and Q(10,y)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.

Q10

Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4).

← Back to Coordinate Geometry