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Coordinate Geometry
EXERCISE 7.1
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

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

Step-by-Step Solution

Key Concept: Use the Distance Formula derived from Pythagoras theorem: For points \((x_1,y_1)\) and \((x_2,y_2)\), the distance \(d\) is \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).
1. Identify the coordinates
- First pair of points: \(P_1(0,0)\) and \(P_2(36,15)\).
- Towns A and B (as given in Section 7.2): \(A(2,3)\) and \(B(10,-5)\).

2. Apply the distance formula
- For \(P_1\) and \(P_2\):
$$d_{P_1P_2}=\sqrt{(36-0)^2+(15-0)^2}$$
- For towns A and B:
$$d_{AB}=\sqrt{(10-2)^2+(-5-3)^2}$$

3. Calculate the squares
- \((36-0)^2 = 36^2 = 1296\)
- \((15-0)^2 = 15^2 = 225\)
- Sum: \(1296+225 = 1521\)
- \((10-2)^2 = 8^2 = 64\)
- \((-5-3)^2 = (-8)^2 = 64\)
- Sum: \(64+64 = 128\)

4. Take the square roots
- \(d_{P_1P_2}=\sqrt{1521}=39\) units.
- \(d_{AB}=\sqrt{128}=\sqrt{64\times2}=8\sqrt{2}\) units.

5. State the results
- Distance between (0,0) and (36,15) = 39 units.
- Distance between towns A and B = $8\sqrt{2}$ units.

Correct Answer: Distance between (0,0) and (36,15) = 39 units. Distance between towns A and B = $8\sqrt{2}$ units.
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