Find the value of x so the distance between the points A(3,5) and B(x,8) is 5 units.
5 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
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.
sciencesolve | Teacher | (Level 3) Educator Emeritus
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).`
hala718 | High School Teacher | (Level 1) Educator Emeritus
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'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.
neela | High School Teacher | (Level 3) Valedictorian
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.