NCERT Class X Chapter 7: Coordinate Geometry Exercise 7.1 Question1(ii)

NCERT Class X Chapter 7: Coordinate Geometry

Question:

Find the distance between the points (–5, 7) and (–1, 3).

Given:

Points \(A(-5, 7)\) and \(B(-1, 3)\).

To Find:

The distance between points \(A\) and \(B\).

Formula:

Distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
$$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Solution:

Step 1: Assign the coordinates.

$$ x_1 = -5,\quad y_1 = 7,\quad x_2 = -1,\quad y_2 = 3 $$

Step 2: Substitute the values into the distance formula.

$$ AB = \sqrt{(-1 - (-5))^2 + (3 - 7)^2} $$

Step 3: Simplify the expressions inside the square root.

$$ AB = \sqrt{(4)^2 + (-4)^2} $$

Step 4: Calculate the squares and add.

$$ AB = \sqrt{16 + 16} $$

Step 5: Add the values inside the root.

$$ AB = \sqrt{32} $$

Step 6: Express 32 as \(16 \times 2\) and simplify.

$$ AB = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4\sqrt{2} $$

Result:

The distance between the points (–5, 7) and (–1, 3) is $$4\sqrt{2}$$ units.

Next question solution:

NCERT Class X Chapter 7: Coordinate Geometry Exercise 7.1 Question 1(iii)

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