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

Question

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

Accepted Answer

∻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.⚹◊What is the equation representing the distance between the two points $(2, n)$ and $(-1,1)$ using the distance formula $d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}$?⍭ Generate me a similar question◊

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