Topic: Distance in Coordinate Plane

Determine the value(s) of n if...
J(n, n+2), K(3n, n-1), JK=5

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To find the value(s) of n in this scenario, we need to determine the coordinates of points J and K.

Point J is represented by J(n, n + 2), where the x-coordinate is n and the y-coordinate is n + 2.

Point K is represented by K(3n, n - 1), where the x-coordinate is 3n and the y-coordinate is n - 1.

The distance between points J and K is given as JK = 5.

To find the value(s) of n, we can use the distance formula to calculate the distance between these two points and set it equal to 5. The distance formula is:

√((x2 - x1)^2 + (y2 - y1)^2) = JK

Substituting the given coordinates:

√((3n - n)^2 + (n - 1 - (n + 2))^2) = 5

Simplifying:

√((2n)^2 + (n - 1 - n - 2)^2) = 5

√((4n^2) + (-3)^2) = 5

√(16n^2 + 9) = 5

Squaring both sides to eliminate the square root:

16n^2 + 9 = 5^2

16n^2 + 9 = 25

16n^2 = 25 - 9

16n^2 = 16

Dividing both sides by 16:

n^2 = 1

Taking the square root of both sides:

n = ±1

Therefore, the value(s) of n that satisfy the condition JK = 5 are n = 1 and n = -1.

To find the value(s) of n, we need to consider the distance between points J and K on the coordinate plane.

The distance formula between two points (x₁, y₁) and (x₂, y₂) is given by:
d = √((x₂ - x₁)² + (y₂ - y₁)²)

Let's use this formula to find the distance between J and K, and equate it to 5:

Step 1: Determine the coordinates of J and K
J(n, n+2)
K(3n, n-1)

Step 2: Calculate the distance between J and K
d = √((3n - n)² + ((n-1) - (n+2))²)
= √((2n)² + (-3)²)
= √(4n² + 9)

Step 3: Set the distance equal to 5 and solve for n
√(4n² + 9) = 5

To solve this equation, we square both sides:

4n² + 9 = 5²
4n² + 9 = 25

Now subtract 9 from both sides:

4n² = 25 - 9
4n² = 16

Divide both sides by 4:

n² = 4

Taking the square root of both sides:

n = ±2

Therefore, the value(s) of n that satisfy the given conditions are n = 2 and n = -2.