The problem provides the length of the segment `AB = 5` , hence, using Pythagorean theorem yields:

`AB^2 = (3 - x)^2 + (5 - 8)^2 `

`5^2 = 9 - 6x + x^2 + 9 => 25 - 18 = x^2 - 6x`

`x^2 - 6x - 7 = 0`

You may use the factorization to evaluate the solutions to quadratic equation, such that:

`x^2 - 7x + x - 7 = 0 => (x^2 - 7x) + (x - 7) = 0`

`x(x - 7) + (x - 7) = 0 => (x - 7)(x + 1) = 0 => {(x - 7 = 0),(x + 1 = 0):} => {(x = 7),(x = -1):} `

**Hence, evaluating the missing coordinate x yields that there exists two points whose y coordinate is `y = 8` , such that:` x = 7, B(7,8)` and **`x = -1, B (-1,8).`

Given the points : A(3, 5) and B(x, 8) such that the distance is 5 units:

We will use the distance formula :

AB = sqrt[( xB-xA)^2 + (yB-yA)^2 ]

= sqrt[ ( x -3)^2 + ( 8 - 5)^2]

= sqrt[(x^2 - 6x + 9 + 3^2)

= sqrt(x^2 - 6x + 9 + 9)

= sqrt(x^2 - 6x + 18)

But we know that AB = 5

==> sqrt(x^2 - 6x + 18) = 5

Square both sides:

==> x^2 - 6x + 18 = 25

==> x^2 - 6x + 18 -25 = 0

==> x^2 - 6x - 7 = 0

==> We will factor:

==>(x-7) ( x+ 1) = 0

==> x1= 7

==> x2 = -1

Then we have two solutions:

**B ( 7,8) OR B(-1, 8)**

We have to find x if the distance between (3,5 ) and ( x,8) is 5.

Now the distance between ( x1, y1) and ( x2, y2) is given by sqrt[(x1-x2)^2+(y1-y2)^]

Here x1 = 3, y1 = 5, x2 = x and y2 = 8

=> sqrt [ (3-x)^2 + (5-8)^2] = 5

take square of both the sides

=> (3-x)^2 + (5-8)^2 = 25

=> (3 -x)^2 + 9 = 25

subtract 9 from both the sides.

=> (3 - x)^2 = 25 - 9

=> (3 - x)^2 = 16

take the square root

=> 3 -x = 4 or 3-x = -4

=> x = -1 or x = 7

**Therefore x can be either -1 or 7.**

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