Question

Segment FG begins at point F(-2, 4) and ends at point G (-2, -3). Segment FG is translated by (x, y) → (x – 3, y + 2) and then reflected across the y-axis to form segment F'G'. How long is segment F'G'?

Answer 1

Answer:

The length of the segment F'G' is 7.

Step-by-step explanation:

From Linear Algebra we define reflection across the y-axis as follows:

$(x',y')=(-x, y)$, $\forall\, x, y\in \mathbb{R}$ (Eq. 1)

In addition, we get this translation formula from the statement of the problem:

$(x',y') =(x-3,y+2)$, $\forall \,x,y\in \mathbb{R}$ (Eq. 2)

Where:

$(x, y)$ - Original point, dimensionless.

$(x', y')$ - Transformed point, dimensionless.

If we know that $F(x,y) = (-2, 4)$ and $G(x,y) = (-2,-3)$, then we proceed to make all needed operations:

Translation

$F''(x,y) = (-2-3,4+2)$

$F''(x,y) = (-5,6)$

$G''(x,y) = (-2-3,-3+2)$

$G''(x,y) = (-5,-1)$

Reflection

$F'(x,y) = (5, 6)$

$G'(x,y) = (5,-1)$

Lastly, we calculate the length of the segment F'G' by Pythagorean Theorem:

$F'G' = \sqrt{(5-5)^{2}+[(-1)-6]^{2}}$

$F'G' = 7$

The length of the segment F'G' is 7.