NCERT Solutions
Class 10 - Mathematics - Chapter 7: COORDINATE GEOMETRY - Exercise 7.1
Question 1

Question. 1

Find the distance between the following pairs of points:

  1. (2, 3), (4, 1)
  2. (−5, 7), (−1, 3)
  3. (a, b), (−a, −b)

Answer:

(1) \(2\sqrt{2}\)

(2) \(4\sqrt{2}\)

(3) \(2\sqrt{a^2 + b^2}\)

Video Explanation:

Detailed Answer with Explanation:

(1)

The given points are \((2, 3)\) and \((4, 1)\).

We use the distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

Substituting the values: \( d = \sqrt{(4 - 2)^2 + (1 - 3)^2} \).

This becomes: \( d = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} \).

Thus, the distance is \( d = 2\sqrt{2} \).

(2)

The given points are \((-5, 7)\) and \((-1, 3)\).

Apply the distance formula: \( d = \sqrt{(-1 + 5)^2 + (3 - 7)^2} \).

Simplifying inside the brackets gives: \( d = \sqrt{4^2 + (-4)^2} \).

Then: \( d = \sqrt{16 + 16} = \sqrt{32} \).

So, the distance is \( d = 4\sqrt{2} \).

(3)

The given points are \((a, b)\) and \((-a, -b)\).

Using the formula: \( d = \sqrt{(-a - a)^2 + (-b - b)^2} \).

This simplifies to: \( d = \sqrt{(-2a)^2 + (-2b)^2} \).

Compute squares: \( (-2a)^2 = 4a^2 \) and \( (-2b)^2 = 4b^2 \).

Thus: \( d = \sqrt{4a^2 + 4b^2} = \sqrt{4(a^2 + b^2)} \).

Therefore, the distance is \( d = 2\sqrt{a^2 + b^2} \).

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