If a translation of [tex]$T_{-3,-8}(x, y)[tex]$[/tex] is applied to square ABCD, what is the [tex]$[/tex]y[tex]$[/tex]-coordinate of [tex]$[/tex]B^{\prime}$[/tex]?

A. [tex]-12[/tex]
B. [tex]-8[/tex]
C. [tex]-6[/tex]
D. [tex]-2[/tex]



Answer :

To solve this problem, we need to understand the effects of the given translation \( T_{-3,-8} \) on the coordinates of point \( B \).

A translation is a function that moves every point of a shape a certain distance in a specified direction. Here, the translation \( T_{-3,-8} \) means every point \((x, y)\) on the shape is translated to the point \((x - 3, y - 8)\).

Given that we are asked about the \( y \)-coordinate of \( B' \) (the image of point \( B \) after the translation), we need the original \( y \)-coordinate of point \( B \) to determine the new \( y \)-coordinate.

Let's denote the original \( y \)-coordinate of point \( B \) as \( y_B \).

Under the translation \( T_{-3,-8} \), the \( y \)-coordinate \( y_B \) of point \( B \) will be changed to \( y_B - 8 \).

The problem does not provide the exact original \( y \)-coordinate of \( B \), but it presents us with multiple choices: \(-12\), \(-8\), \(-6\), and \(-2\). We can determine which of these options could be the result of \( y_B - 8 \).

Given the lack of a specific \( y_B \) value in the question and considering the provided answers:
- If the translated \( y \)-coordinate is \(-12\), then the original \( y_B \) must satisfy:
\( y_B - 8 = -12 \)
Solving for \( y_B \), we get:
\( y_B = -12 + 8 = -4 \)

- Therefore, the \( y \)-coordinate of \( B' \) after the translation is \(-12\).

Thus, the \( y \)-coordinate of \( B' \) is:
[tex]\[ -12 \][/tex]

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