Find the point on x axis which is equidistant from 2, 25 and 22 ,9
Answers
Given : point on x axis is equidistant from 2, 25 and 22 ,9
To Find : The point
Solution:
Point on x axis = ( x , 0)
Equidistant from points ( 2 , 25) & ( 22, 9)
Distance between ( x , 0) and ( 2 , 25)
= √(x - 2)² + (0 - 25)²
= √(x - 2)² +25²
Distance between ( x , 0) and ( 22 , 9)
= √(x - 22)² + (0 - 9)²
= √(x - 22)² +9²
√(x - 2)² +25² = √(x - 22)² +9²
=> (x - 2)² +25² = (x - 22)² +9²
=> x² - 4x + 4 + 625 = x² - 44x + 484 + 81
=> 40x + 629 = 565
=> 40x = -64
=> x = - 64/40
=> x = -8/5
=> x = -1.6
point on x axis which is equidistant from 2, 25 and 22 ,9 is ( - 1.6 , 0)
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