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CBSE Class 10 Basic Math 2026 All Sets Solved Paper

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Question : 3 of 20
Marks: +1, -0
The distance between the points (−1,22) and (2,2)(-1, 2\sqrt{2})\, \text{and}\, (2, \sqrt{2}) is:
Solution:  
Given:
A(x1,y1)=(−1,22)A (x_1, y_1) = (-1, 2\sqrt{2})
B(x2,y2)=(2,2)B(x_2, y_2) = (2, \sqrt{2})
The distance between two given point (AB) = (x2−x1)2+(y2−y1)2\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}
=(2−(−1))2+(2−22)2= \sqrt{(2 - (-1))^2 + (\sqrt{2} - 2\sqrt{2})^2}
=(3)2+(−2)2= \sqrt{(3)^2 + (-\sqrt{2})^2}
=9+2= \sqrt{9+2}
=11= \sqrt{11}
The distance between the points is 11\sqrt{11} units
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