Homework Help

Find the value of x so the distance between the points A(3,5) and B(x,8) is 5 units.

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ulichh | Student, College Freshman | eNoter

Posted November 28, 2010 at 12:20 AM via web

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Find the value of x so the distance between the points A(3,5) and B(x,8) is 5 units.

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giorgiana1976 | College Teacher | Valedictorian

Posted November 28, 2010 at 12:20 AM (Answer #1)

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We'll write the distance formula:

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

We'll substitute the coordinates for A and B in the formula:

A(3,5) and B(x,8)

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

We'll substitute the distance by 5:

5 =  sqrt[(x - 3)^2 + (3)^2]

We'll raise to square both sides:

25 = (x-3)^2 + 9

We'll subtract 9 both sides:

25 - 9 = (x-3)^2

16 = (x-3)^2

We'll extract the square root:

x-3 = +4

x = 7

or

x - 3 = -4

x = -1

Both values of x are valid solutions.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted November 28, 2010 at 12:21 AM (Answer #2)

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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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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted November 28, 2010 at 12:22 AM (Answer #3)

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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)

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neela | High School Teacher | Valedictorian

Posted November 28, 2010 at 1:16 AM (Answer #4)

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The distance d between the points (x1,y1) and (x2,y2) is given by:

 d^2 = (x2-x1)^2+(y2-y1)^2.

Therefore the distance between the points A(3,5) and B(x,8) is given by:

d^2 = (x-3)^2+(8-5)^2.

Since  the actual distance d between A and B is 5, we get:

5^2 = (x-3)^2 +3^2.

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

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

16 = (x-3)^2.

Taking the square root, we get:

(x-3) = sqrt16  = 4, Or x-3 = -sqrt16 = -4.

Therefore x= 4+3 = 7. Or x= -4+3 = -1.

Therefore  x coordinate can be 7 or -1.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted March 11, 2013 at 4:18 PM (Answer #3)

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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).`

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