Point $J$ is located at $(-4,4)$ . $\newline$ Select all of the following that are $4$ units from point $J$ . $\newline$ Choose all answers that apply: $\newline$ (A) $x$ -axis $\newline$ (B) $(-4,8)$ $\newline$ (C) $y$ -axis

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Grade 8

Mean, median, mode, and range: find the missing number

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Q. Point

J

is located at

(-4,4)

\newline

Select all of the following that are

4

units from point

J

\newline

Choose all answers that apply:

\newline

(A)

x

-axis

\newline

(B)

(-4,8)

\newline

(C)

y

-axis

Distance Formula Application: To determine which options are $4$ units away from point $J$ , we need to consider the distance formula in a $2D$ coordinate system, which is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ , where $(x_1, y_1)$ is the original point and $(x_2, y_2)$ is the point we're measuring the distance to. We will apply this formula to each option to see if the distance is $4$ units.
Option A Calculation: For option A (x-axis), we need to find a point on the x-axis that is $4$ units away from $J$ . Since $J$ is at $(-4,4)$ , a point on the x-axis that is $4$ units away would have to be at $(-4,0)$ because the $y$ -coordinate would be $0$ (on the x-axis) and $4$ units away in the $y$ -direction. Let's calculate the distance: $\newline$ $d = \sqrt{(-4 - (-4))^2 + (0 - 4)^2}$ $\newline$ $d = \sqrt{(0)^2 + (-4)^2}$ $\newline$ $d = \sqrt{0 + 16}$ $\newline$ $d = \sqrt{16}$ $\newline$ $d = 4$ $\newline$ The x-axis is indeed $4$ units away from point $J$ at the specific point $(-4,0)$ .
Option B Calculation: For option B $(-4,8)$ , we will use the distance formula to find the distance from $J$ to this point: $\newline$ $d = \sqrt{(-4 - (-4))^2 + (8 - 4)^2}$ $\newline$ $d = \sqrt{(0)^2 + (4)^2}$ $\newline$ $d = \sqrt{0 + 16}$ $\newline$ $d = \sqrt{16}$ $\newline$ $d = 4$ $\newline$ The point $(-4,8)$ is also $4$ units away from point $J$ .
Option C Calculation: For option C (y-axis), we need to find a point on the y-axis that is $4$ units away from $J$ . Since $J$ is at $(-4,4)$ , a point on the y-axis that is $4$ units away would have to be at $(0,4)$ because the $x$ -coordinate would be $0$ (on the y-axis) and it would not be $4$ units away in either direction since the $y$ -coordinate is the same as $J$ 's. Therefore, the y-axis is not $4$ units away from point $J$ at any specific point.