Question

general directions: for multiple - choice items, solve each problem and choose the best answer from the choices provided. for open - ended items, demonstrate your knowledge and skills by giving a clear, concise solution to each problem. be sure to include relevant calculations and justify your answer. answers must be simplified. 1. the vertices of triangle ktz are graphed below. k(-4,4) z(5,1) t(-1,-2) part a. calculate the length of each side of triangle ktz using the distance formula. 1) $overline{kz}=$ 2) $overline{tz}=$ 3) $overline{kt}=$ part b. is the triangle ktz equilateral? how do you know? 2. the coordinate plane below shows the graph of line g. which of the equations below describe the graph of a line parallel to line g, perpendicular to line g, or neither parallel nor parallel to line g. write each equation into the appropriate box. parallel perpendicular neither $4x + y = 0$ $y=-\frac{1}{4}x - 3$ $y = 4x+6$ $y=\frac{1}{4}x - 6$ $x + 4y = 0$ $4x - y = 0$

Explanation:

Step1: Recall distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Calculate $\overline{KZ}$

For points $K(-4,4)$ and $Z(5,1)$, we have $x_1=-4,y_1 = 4,x_2=5,y_2 = 1$.
\[

$$\begin{align*} \overline{KZ}&=\sqrt{(5-(-4))^2+(1 - 4)^2}\\ &=\sqrt{(5 + 4)^2+(1 - 4)^2}\\ &=\sqrt{9^2+(-3)^2}\\ &=\sqrt{81 + 9}\\ &=\sqrt{90}\\ &=3\sqrt{10} \end{align*}$$

\]

Step3: Calculate $\overline{TZ}$

For points $T(-1,-2)$ and $Z(5,1)$, we have $x_1=-1,y_1=-2,x_2 = 5,y_2=1$.
\[

$$\begin{align*} \overline{TZ}&=\sqrt{(5-(-1))^2+(1-(-2))^2}\\ &=\sqrt{(5 + 1)^2+(1 + 2)^2}\\ &=\sqrt{6^2+3^2}\\ &=\sqrt{36+9}\\ &=\sqrt{45}\\ &=3\sqrt{5} \end{align*}$$

\]

Step4: Calculate $\overline{KT}$

For points $K(-4,4)$ and $T(-1,-2)$, we have $x_1=-4,y_1 = 4,x_2=-1,y_2=-2$.
\[

$$\begin{align*} \overline{KT}&=\sqrt{(-1-(-4))^2+(-2 - 4)^2}\\ &=\sqrt{(-1 + 4)^2+(-2-4)^2}\\ &=\sqrt{3^2+(-6)^2}\\ &=\sqrt{9 + 36}\\ &=\sqrt{45}\\ &=3\sqrt{5} \end{align*}$$

\]

Step5: Determine if triangle is equilateral

An equilateral triangle has all side - lengths equal. Since $\overline{KZ}=3\sqrt{10}$, $\overline{TZ}=3\sqrt{5}$, and $\overline{KT}=3\sqrt{5}$, and $3\sqrt{10}
eq3\sqrt{5}$, triangle $KTZ$ is not equilateral.

Step6: Find slope of line $g$

Let's assume two points on line $g$ are $(0,6)$ and $(8,2)$. The slope $m_g=\frac{2 - 6}{8-0}=\frac{-4}{8}=-\frac{1}{2}$.
For a line $y=mx + b$, parallel lines have the same slope and perpendicular lines have slopes that are negative reciprocals.

• For the line $4x + y=0$ (or $y=-4x$), slope $m=-4$. It is neither parallel nor perpendicular.
• For the line $y=-\frac{1}{4}x-3$, slope $m =-\frac{1}{4}$. It is neither parallel nor perpendicular.
• For the line $y = 4x+6$, slope $m = 4$. It is neither parallel nor perpendicular.
• For the line $y=\frac{1}{4}x-6$, slope $m=\frac{1}{4}$. It is neither parallel nor perpendicular.
• For the line $x + 4y=0$ (or $y=-\frac{1}{4}x$), slope $m=-\frac{1}{4}$. It is neither parallel nor perpendicular.
• For the line $4x-y=0$ (or $y = 4x$), slope $m = 4$. It is neither parallel nor perpendicular.

Answer:

1.

1. $\overline{KZ}=3\sqrt{10}$
2. $\overline{TZ}=3\sqrt{5}$
3. $\overline{KT}=3\sqrt{5}$

2.
Parallel: None
Perpendicular: None
Neither: $4x + y = 0$, $y=-\frac{1}{4}x-3$, $y = 4x+6$, $y=\frac{1}{4}x-6$, $x + 4y=0$, $4x-y=0$