Question

rotating $\triangle pqr$ by $180^{\circ}$ clockwise about the origin, we get its image $\triangle pqr$ as shown below. note that $\triangle pqr$ has vertices $p(-2, -6)$, $q(-5, -8)$, and $r(-7, -3)$. also, note that $\triangle pqr$ has vertices $p(2, 6)$, $q(5, 8)$, and $r(7, 3)$. complete the following. (a) suppose each pair of segments below have the same length. find each length. give exact answers (not decimal approximations). $cp = cp = \square$ units $cq = cq = \square$ units $cr = cr = \square$ units

Explanation:

Step1: Recall the distance formula

The distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is given by \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Here, point \(C\) is the origin \((0,0)\), and we need to find the distance from \(C\) to \(P\), \(C\) to \(Q\), and \(C\) to \(R\) (since rotation about the origin preserves distance, so \(CP = CP'\), \(CQ=CQ'\), \(CR = CR'\)).

Step2: Calculate \(CP\) (distance from \(C(0,0)\) to \(P(-2,-6)\))

Using the distance formula:
\[
CP=\sqrt{(-2 - 0)^2+(-6 - 0)^2}=\sqrt{(-2)^2+(-6)^2}=\sqrt{4 + 36}=\sqrt{40}=\sqrt{4\times10}=2\sqrt{10}
\]

Step3: Calculate \(CQ\) (distance from \(C(0,0)\) to \(Q(-5,-8)\))

Using the distance formula:
\[
CQ=\sqrt{(-5 - 0)^2+(-8 - 0)^2}=\sqrt{(-5)^2+(-8)^2}=\sqrt{25 + 64}=\sqrt{89}
\]

Step4: Calculate \(CR\) (distance from \(C(0,0)\) to \(R(-7,-3)\))

Using the distance formula:
\[
CR=\sqrt{(-7 - 0)^2+(-3 - 0)^2}=\sqrt{(-7)^2+(-3)^2}=\sqrt{49+9}=\sqrt{58}
\]

Answer:

\(CP = CP'=\boldsymbol{2\sqrt{10}}\) units, \(CQ = CQ'=\boldsymbol{\sqrt{89}}\) units, \(CR = CR'=\boldsymbol{\sqrt{58}}\) units