Question

that \\( \triangle x y z \\) has vertices \\( x(-2,-1) \\), \\( z(-1,-6) \\). also, note that \\( \triangle x^{prime} y^{prime} z^{prime} \\) has vertices \\( x^{prime}(3,3), y^{prime} \\) and \\( z^{prime}(4,-2) \\). complete the following. (a) find each length. give exact answers (not decimal approximations). \\ \

$$\begin{array}{l} x x^{prime}=\\text { units } \\\\ y y^{prime}=\\text { units } \\\\ z z^{prime}=\\text { units } \\end{array}$$

\\ (b) find each slope. give exact answers (not decimal appros). \\ \

$$\begin{array}{l} \\text { slope of } \\overline{x x^{prime}}= \\\\ \\text { slope of } \\overline{y y^{prime}}= \\\\ \\text { slope of } \\overline{z z^{prime}}= \\end{array}$$

\\

Explanation:

Step1: Find the length of \(XX'\)

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For \(X(-2,-1)\) and \(X'(3,3)\), \(x_1=-2,y_1 = - 1,x_2=3,y_2 = 3\).
\(XX'=\sqrt{(3-(-2))^2+(3 - (-1))^2}=\sqrt{(3 + 2)^2+(3 + 1)^2}=\sqrt{25 + 16}=\sqrt{41}\)

Step2: Find the length of \(YY'\)

For \(Y(-5, - 4)\) and \(Y'(-1,2)\), \(x_1=-5,y_1=-4,x_2=-1,y_2 = 2\).
\(YY'=\sqrt{(-1-(-5))^2+(2-(-4))^2}=\sqrt{(-1 + 5)^2+(2 + 4)^2}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}\)

Step3: Find the length of \(ZZ'\)

For \(Z(-1,-6)\) and \(Z'(4,-2)\), \(x_1=-1,y_1=-6,x_2=4,y_2=-2\).
\(ZZ'=\sqrt{(4-(-1))^2+(-2-(-6))^2}=\sqrt{(4 + 1)^2+(-2 + 6)^2}=\sqrt{25 + 16}=\sqrt{41}\)

Step4: Find the slope of \(\overline{XX'}\)

Use the slope formula \(m=\frac{y_2-y_1}{x_2-x_1}\). For \(X(-2,-1)\) and \(X'(3,3)\), \(m_{XX'}=\frac{3-(-1)}{3-(-2)}=\frac{3 + 1}{3+2}=\frac{4}{5}\)

Step5: Find the slope of \(\overline{YY'}\)

For \(Y(-5,-4)\) and \(Y'(-1,2)\), \(m_{YY'}=\frac{2-(-4)}{-1-(-5)}=\frac{2 + 4}{-1 + 5}=\frac{6}{4}=\frac{3}{2}\)

Step6: Find the slope of \(\overline{ZZ'}\)

For \(Z(-1,-6)\) and \(Z'(4,-2)\), \(m_{ZZ'}=\frac{-2-(-6)}{4-(-1)}=\frac{-2 + 6}{4+1}=\frac{4}{5}\)

Answer:

(a) \(XX'=\sqrt{41}\) units, \(YY' = 2\sqrt{13}\) units, \(ZZ'=\sqrt{41}\) units
(b) Slope of \(\overline{XX'}=\frac{4}{5}\), Slope of \(\overline{YY'}=\frac{3}{2}\), Slope of \(\overline{ZZ'}=\frac{4}{5}\)