Answer:
x = - 5
Step-by-step explanation:
[tex]Let \ (x _ 1 , y _ 1 ) \ and \ (x _ 2 , y _ 2 ) \ be \ the \ points. \\\\The \ distance \ between \ the \ points \ be ,\ d = \sqrt{(x_2 - x_1)^2 + ( y _ 2 - y_1)^2}[/tex]
Given : d = 10 units
And the points are ( x , - 4) and ( 3 , 2 ).
Find x
[tex]d = \sqrt{( 3 - x)^2 + ( -4 - 2)^2} \\\\10 = \sqrt{( 3 - x)^2 + ( -6)^2} \\\\10^2 = [ \ \sqrt{( 3 - x)^2 + 36} \ ]^2 \ \ \ \ \ \ \ \ \ [ \ squaring \ both \ sides \ ] \\\\100 = ( 3 - x )^2 + 36\\\\100 - 36 = ( 3 - x )^ 2\\\\( 3 - x ) = \sqrt{64}\\\\3 - x = \pm 8\\\\3 - x = 8 \ and \ 3 - x = - 8\\\\-x = 8 - 3 \ and \ -x = - 8 - 3\\\\-x = 5 \ and \ -x = - 11\\\\x = - 5 \ and \ x = 11\\\\[/tex]
Check which value of x satisfies the distance between the points.
x = 11
[tex]d = \sqrt{(3-11)^2 + (-2--4)^2} = \sqrt{(-8)^2 + (-2+4)^2}= \sqrt{64+4} = \sqrt {68} \ units[/tex]
does not satisfy.
x = - 5:
[tex]d = \sqrt{ (3 -- 5)^2 + ( - 4 - 2)^2} = \sqrt{8^2 + 6^2} = \sqrt{100} =10 \ units[/tex]
Therefore , x = - 5
