Question

5.) find the other endpoint of the line segment with the given endpoint and midpoint.
endpoint: (6,5) midpoint: (4,2)
6.) reflect the triangle first over the y - axis, translate 3 units down, then dilate the image by a scale factor of 1/2. next, draw the resulting image with the coordinates of the triangle written below the graph.
7.) cannot use linear pairs or vertical angles. line gb is || to hc
a.) solve for x (must use proof sheet)
b.) given your solution from m∠afb find the m∠aec
c.) given your solution from m∠hed find the m∠dfg
d.) given your solution from m∠afb find the m∠dec
8.) given m∠utv = 25°
a.) find the m∠vtx (must use proof sheet)
b.) find the m∠xyz given m∠uty = 27b + 20 solve for b

Explanation:

5.

Step1: Recall mid - point formula

The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $M=(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Let the given endpoint be $(x_1,y_1)=(6,5)$ and the mid - point be $(M_x,M_y)=(4,2)$. We want to find $(x_2,y_2)$.
For the x - coordinate:
$M_x=\frac{x_1 + x_2}{2}$, so $4=\frac{6 + x_2}{2}$.

Step2: Solve for $x_2$

Multiply both sides of the equation $4=\frac{6 + x_2}{2}$ by 2: $4\times2=6 + x_2$. Then $8=6 + x_2$. Subtract 6 from both sides: $x_2=8 - 6=2$.
For the y - coordinate:
$M_y=\frac{y_1 + y_2}{2}$, so $2=\frac{5 + y_2}{2}$.

Step3: Solve for $y_2$

Multiply both sides of the equation $2=\frac{5 + y_2}{2}$ by 2: $2\times2=5 + y_2$. Then $4=5 + y_2$. Subtract 5 from both sides: $y_2=4 - 5=-1$.

Step1: Rearrange the equation

Subtract $25x$ from both sides of the equation $25x+5=26x + 2$: $5=x + 2$.

Step2: Solve for $x$

Subtract 2 from both sides: $x=5 - 2=3$.

Answer:

The other endpoint is $(2,-1)$

6.

Let's assume the vertices of the original triangle are $(x_1,y_1),(x_2,y_2),(x_3,y_3)$.
Reflection over the y - axis:
The rule for reflecting a point $(x,y)$ over the y - axis is $(x,y)\to(-x,y)$. So the new vertices are $(-x_1,y_1),(-x_2,y_2),(-x_3,y_3)$.
Translation 3 units down:
The rule for translating a point $(x,y)$ 3 units down is $(x,y)\to(x,y - 3)$. So the vertices become $(-x_1,y_1-3),(-x_2,y_2 - 3),(-x_3,y_3 - 3)$.
Dilation by a scale factor of $\frac{1}{2}$:
The rule for dilating a point $(x,y)$ by a scale factor $k=\frac{1}{2}$ centered at the origin is $(x,y)\to(kx,ky)$. So the final vertices are $(-\frac{1}{2}x_1,\frac{y_1-3}{2}),(-\frac{1}{2}x_2,\frac{y_2 - 3}{2}),(-\frac{1}{2}x_3,\frac{y_3 - 3}{2})$.

7a.

Since line $GB\parallel HC$, corresponding angles are equal. $\angle AFB$ and $\angle HED$ are corresponding angles. So $25x+5=26x + 2$.