Find all possible values of x for which the distance between the points A(x, -1) and B(5, 3)is 5 units?

Question

Answer

Root under (x2 - x1)^2 + ( y2 - y1)^2 = 5
(5 - x )^2 + { 3 - ( - 1) }^2 = (5)^2
25 + x^2 - 10 x + (4)^2 = 25
25 + x^2 - 10 x + 16 = 25
x^2 - 10 x + 41 - 25 = 0
x^2 - 10 x + 16 = 0
x^2 - 8x - 2 x + 16 = 0
x ( x - 8 ) - 2 ( x - 8 ) = 0
( x - 2 ) ( x - 8 ) = 0
x - 2 = 0
x = 2
x - 8 = 0
x = 8
Possible values of x = 2 & 8

Answer

.....x=2 and 8

Answer

Answer

Answer

Answer

Ya... But Can we Chat..?

Answer

X=2 and 8

Answer

Introduction:
We are given two points A(x, -1) and B(5, 3). We need to find all possible values of x for which the distance between these two points is 5 units.

Formula for distance between two points:
To find the distance between two points (x1, y1) and (x2, y2), we use the distance formula:
distance = √((x2 - x1)^2 + (y2 - y1)^2)

Deriving the equation:
Let's use the distance formula to find the distance between points A(x, -1) and B(5, 3):
distance = √((5 - x)^2 + (3 - (-1))^2)
distance = √((5 - x)^2 + (4)^2)

We know that the distance between the points A and B is 5 units. So we can set up the equation:
√((5 - x)^2 + (4)^2) = 5

Solving the equation:
To solve this equation, we need to square both sides to eliminate the square root:
((5 - x)^2 + 4^2) = 5^2
(5 - x)^2 + 16 = 25
(5 - x)^2 = 25 - 16
(5 - x)^2 = 9

Taking the square root of both sides, we have:
5 - x = ±√9
5 - x = ±3

Case 1: 5 - x = 3
Solving for x in this case:
5 - x = 3
x = 5 - 3
x = 2

Case 2: 5 - x = -3
Solving for x in this case:
5 - x = -3
x = 5 + 3
x = 8

Conclusion:
The possible values of x for which the distance between the points A(x, -1) and B(5, 3) is 5 units are x = 2 and x = 8.

Answer

8,2

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