Question

Math

Posted 4 months ago

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possible integer values for $n$ ?
A point $(2, n)$ is 5 units away from the point $(-1,1)$. What is the product of all
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Answer from Sia

Posted 4 months ago

Solution by Steps

step 1

To find the possible integer values for $n$, we use the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$, which is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

step 2

The distance between the points $(2, n)$ and $(-1, 1)$ is given to be 5 units. So, we set up the equation $(2 - (-1))^2 + (n - 1)^2 = 5^2$

step 3

Simplifying the equation, we get $3^2 + (n - 1)^2 = 25$

step 4

Solving for $n$, we subtract $3^2$ from both sides to get $(n - 1)^2 = 16$

step 5

Taking the square root of both sides, we find $n - 1 = \pm4$

step 6

Solving for $n$, we get $n = 5$ and $n = -3$

Answer

The possible integer values for $n$ are 5 and -3.

Key Concept

Distance Formula and Solving Quadratic Equations

Explanation

We used the distance formula to set up an equation and then solved the resulting quadratic equation to find the integer values of $n$.

Solution by Steps

step 1

To find the product of all integer solutions for $n$, we multiply the two solutions we found

step 2

The solutions are $n = 5$ and $n = -3$

step 3

Multiplying these together, we get $5 \times (-3)$

step 4

The product is $-15$

Answer

The product of all integer solutions for $n$ is -15.

Key Concept

Product of Solutions

Explanation

The product of the solutions is found by multiplying the individual solutions for $n$ together.

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