if the distance between the point (4,k) and (1,0) is 5 , then what can be the possible value of k?
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k=4...................
how?
yes
yup
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HOLA MATE !!!
HOPE THIS HELPS YOU...
Answer :-
We can solve this problem using distance formula,
d = \sqrt{(x1 - x2)² + (y1 - y2)²}
Here , x1 = 4 ,x2 = k ,y1 = 1 ,y2 = 0 ,d = 5units
Substitution and simplification:-
=> 5 = \sqrt{(4 - k)² + (1 - 0)²}
=> 5 = \sqrt{(16 + k² -8k +1)}
=> 5 = \sqrt{(17 + k² - 8k)}
squaring on both sides of the equation,
=> k² - 8k - 8 = 0
using formula , k = [-b ± \sqrt{(b² - 4ac)}]/2a
=> k = [8 ± \sqrt{(64 - 32)}] /2
=> k = [8 ± √32]/2
are the suitable values for 'k'.
THANK YOU FOR THE WONDERFUL QUESTION...
#bebrainly
HOPE THIS HELPS YOU...
Answer :-
We can solve this problem using distance formula,
d = \sqrt{(x1 - x2)² + (y1 - y2)²}
Here , x1 = 4 ,x2 = k ,y1 = 1 ,y2 = 0 ,d = 5units
Substitution and simplification:-
=> 5 = \sqrt{(4 - k)² + (1 - 0)²}
=> 5 = \sqrt{(16 + k² -8k +1)}
=> 5 = \sqrt{(17 + k² - 8k)}
squaring on both sides of the equation,
=> k² - 8k - 8 = 0
using formula , k = [-b ± \sqrt{(b² - 4ac)}]/2a
=> k = [8 ± \sqrt{(64 - 32)}] /2
=> k = [8 ± √32]/2
are the suitable values for 'k'.
THANK YOU FOR THE WONDERFUL QUESTION...
#bebrainly
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