Question

1. a square has coordinates m(1, -2), a(-4, -2), t(-4, 3), h(1, 3). if the square is reflected over the line y = -3, find the coordinates of the image points.
2. a triangle has coordinates c(-6, 8), a(7, 3), t(1, -5). after a reflection, the image points of the same triangle are c(4, 8), a(-9, 3), t(-3, -5). what line was the triangle reflected over?
3. triangle dog was reflected over the line x = 4 resulting in the image triangle with coordinates d(4, 8), o(-9, 3), g(-6, -5). which of the following coordinates would not be on triangle dog? a. (4,4) b. (17,3) c. (4,8) d. (14, -5)

Explanation:

3.

Step1: Recall reflection rule over horizontal line

The rule for reflecting a point $(x,y)$ over the horizontal line $y = k$ is $(x,2k - y)$. Here $k=-3$.

Step2: Reflect point $M(1,-2)$

For $M(1,-2)$, using the formula $(x,2k - y)$ with $x = 1,k=-3,y=-2$, we get $(1,2\times(-3)-(-2))=(1,-6 + 2)=(1,-4)$.

Step3: Reflect point $A(-4,-2)$

For $A(-4,-2)$, $(x,2k - y)$ gives $(-4,2\times(-3)-(-2))=(-4,-6 + 2)=(-4,-4)$.

Step4: Reflect point $T(-4,3)$

For $T(-4,3)$, $(x,2k - y)$ gives $(-4,2\times(-3)-3)=(-4,-6 - 3)=(-4,-9)$.

Step5: Reflect point $H(1,3)$

For $H(1,3)$, $(x,2k - y)$ gives $(1,2\times(-3)-3)=(1,-6 - 3)=(1,-9)$.

Step1: Analyze the change in $x$-coordinates

The mid - point of the line segment joining a point $(x_1,y_1)$ and its image $(x_2,y_2)$ under reflection over a vertical line $x = a$ is given by $a=\frac{x_1 + x_2}{2}$. For point $C(-6,8)$ and $C'(4,8)$, $a=\frac{-6 + 4}{2}=\frac{-2}{2}=-1$. For point $A(7,3)$ and $A'(-9,3)$, $a=\frac{7+( - 9)}{2}=\frac{-2}{2}=-1$. For point $T(1,-5)$ and $T'(-3,-5)$, $a=\frac{1+( - 3)}{2}=\frac{-2}{2}=-1$.

Step1: Recall reflection rule over vertical line

The rule for reflecting a point $(x,y)$ over the vertical line $x = a$ is $(2a - x,y)$. Given the line of reflection $x = 4$.

Step2: Check each option

For option A: If $(x,y)=(4,4)$, then $2a - x=2\times4 - 4=4$.
For option B: If $(x,y)=(17,3)$, then $2a - x=2\times4-17=8 - 17=-9$.
For option C: The point $D'(4,8)$ is the image point. The original point of $D'$ under reflection over $x = 4$ should have $x$ - coordinate such that $2\times4 - x=4$, which gives $x = 4$. But this is the image point, not the pre - image point.
For option D: If $(x,y)=(14,-5)$, then $2a - x=2\times4-14=8 - 14=-6$.

Answer:

$M'(1,-4),A'(-4,-4),T'(-4,-9),H'(1,-9)$

4.