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CBSE Class 10 Math 2024 All India Set 1 Solved Paper

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Question : 10 of 20
Marks: +1, -0
If the distance between the points (3, 0) and (2, y) is 5\sqrt{5} then the value(s) of y is:
Solution:  
Given:-
A = (3, 0)
B = (2, y)
Distance between AB=5AB = \sqrt{5}
The distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2)
Distance=(x2−x1)2+(y2−y1)2\text{Distance} = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
5=(2−3)2+(y−0)2\sqrt{5} = \sqrt{(2-3)^2+(y-0)^2}
(5)2=(−1)2+(y)2(\sqrt{5})^2 = (-1)^2+(y)^2
5=1+y25 = 1+y^2
y2=5−1y^2 = 5-1
y=±4y = \pm \sqrt{4}
y=±2y = \pm 2
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