Question

There are two points of the form (x,-4) that have a distance of 10 units from the point (3,2). Give the x value for one of those points.​

Answer 1

x = - 5

Explanation:

`Let \ (x _ 1 , y _ 1 ) \ and \ (x _ 2 , y _ 2 ) \ be \ the \ points. \\\\The \ distance \ between \ the \ points \ be ,\ d = √((x_2 - x_1)^2 + ( y _ 2 - y_1)^2)`

Given : d = 10 units

And the points are ( x , - 4) and ( 3 , 2 ).

Find x

`d = √(( 3 - x)^2 + ( -4 - 2)^2) \\\\10 = √(( 3 - x)^2 + ( -6)^2) \\\\10^2 = [ \ √(( 3 - x)^2 + 36) \ ]^2 \ \ \ \ \ \ \ \ \ [ \ squaring \ both \ sides \ ] \\\\100 = ( 3 - x )^2 + 36\\\\100 - 36 = ( 3 - x )^ 2\\\\( 3 - x ) = √(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\\\\`

Check which value of x satisfies the distance between the points.

x = 11

`d = √((3-11)^2 + (-2--4)^2) = √((-8)^2 + (-2+4)^2)= √(64+4) = \sqrt {68} \ units`

does not satisfy.

x = - 5:

`d = √( (3 -- 5)^2 + ( - 4 - 2)^2) = √(8^2 + 6^2) = √(100) =10 \ units`

Therefore , x = - 5