Question

what are the steps necessary to show that $overline{jh}congoverline{vu}$?
check all that apply.
$square vu=sqrt{15}$
$square jh = d=sqrt{(-3-(-5))^{2}+(0-(-3))^{2}}$
$square vu = d=sqrt{(5-(2))^{2}+(2 - 4)^{2}}$
$square vw = d=sqrt{(0-(5))^{2}+(1 - 2)^{2}}$
$square jh=sqrt{13}$

Explanation:

Step1: Calculate the length of \(JH\)

Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), for points \(J(-5,-3)\) and \(H(-3,0)\), we have \(JH = d=\sqrt{(-3-(-5))^2+(0 - (-3))^2}=\sqrt{(2)^2+(3)^2}=\sqrt{4 + 9}=\sqrt{13}\)

Step2: Calculate the length of \(VU\)

Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), for points \(V(5,2)\) and \(U(2,4)\), we have \(VU = d=\sqrt{(5 - 2)^2+(2 - 4)^2}=\sqrt{(3)^2+(-2)^2}=\sqrt{9+4}=\sqrt{13}\)

Answer:

Second option (\(JH = d=\sqrt{(-3-(-5))^2+(0 - (-3))^2}\)), third option (\(VU = d=\sqrt{(5 - 2)^2+(2 - 4)^2}\)), fifth option (\(JH=\sqrt{13}\))