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 AA and BB discussed in Section 7.2.

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Solution
Understand the Question
  • The distance between any two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is calculated using the Distance Formula: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
  • This formula is derived directly from the Pythagoras theorem, where the distance is the hypotenuse of a right-angled triangle.
  • In Section 7.2 of NCERT, Town AA is located at the origin (0,0)(0, 0) and Town BB is located 36 km36\text{ km} East and 15 km15\text{ km} North, corresponding to coordinates (36,15)(36, 15). Hence, the same distance calculation applies.

Step 1 · Calculate Distance Between the Given Points

Let the points be P(0,0)P(0, 0) and Q(36,15)Q(36, 15).Diagram 1

Using the distance formula:

PQ=(x2−x1)2+(y2−y1)2=(36−0)2+(15−0)2=362+152=1296+225=1521=39 units\begin{aligned} PQ &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[0.6em] &= \sqrt{(36 - 0)^2 + (15 - 0)^2} \\[0.6em] &= \sqrt{36^2 + 15^2} \\[0.6em] &= \sqrt{1296 + 225} \\[0.6em] &= \sqrt{1521} \\[0.6em] &= 39\text{ units} \end{aligned}

Step 2 · Find Distance Between Towns AA and BB

In Section 7.2, Town AA is taken at the origin (0,0)(0, 0) and Town BB is located at (36,15)(36, 15) (36 km36\text{ km} East and 15 km15\text{ km} North).

Since the coordinates are (0,0)(0, 0) and (36,15)(36, 15): Distance between Town A and Town B=39 km\text{Distance between Town } A \text{ and Town } B = 39\text{ km}

Yes, we can find the distance between the two towns.

Answer

The distance between (0,0)(0, 0) and (36,15)(36, 15) is 39 units39\text{ units}. Yes, the distance between towns AA and BB is 39 km39\text{ km}.

Common Mistakes
  • Forgetting the Square Root: Students often compute (x2−x1)2+(y2−y1)2=1521(x_2 - x_1)^2 + (y_2 - y_1)^2 = 1521 and forget to take the final square root 1521=39\sqrt{1521} = 39.
  • Arithmetic Errors in Large Squares: Squaring 362=129636^2 = 1296 and finding 1521\sqrt{1521} requires careful long division or prime factorisation (1521=32×132=3921521 = 3^2 \times 13^2 = 39^2).
  • Missing Units: Writing purely numeric answers without specifying units\text{units} for coordinates or km\text{km} for the distance between towns.

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 AA and BB 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)(5, -2), (6,4)(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 AA, BB, CC and DD 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 ABCDABCD 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)Q(0, 1) is equidistant from P(5,−3)P(5, -3) and R(x,6)R(x, 6), find the values of xx. Also find the distances QRQR and PRPR.

Q10

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

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