Question

show that a quadrilateral with the given vertices is a parallelogram. find the lengths of the sides of the quadrilateral. pa = □ tp = □ tr = □ ra = □ (simplify your answers. type exact answers, using radicals as needed.)

Explanation:

Step1: <Identify coordinates>

Let \(P(-2,1)\), \(A(2,2)\), \(R(1,-2)\), \(T(-3,-3)\)

Step2: <Use distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) for \(PA\)>

\(PA=\sqrt{(2-(-2))^2+(2 - 1)^2}=\sqrt{(4)^2+(1)^2}=\sqrt{16 + 1}=\sqrt{17}\)

Step3: <Use distance formula for \(TP\)>

\(TP=\sqrt{(-2-(-3))^2+(1-(-3))^2}=\sqrt{(1)^2+(4)^2}=\sqrt{1+16}=\sqrt{17}\)

Step4: <Use distance formula for \(TR\)>

\(TR=\sqrt{(1-(-3))^2+(-2-(-3))^2}=\sqrt{(4)^2+(1)^2}=\sqrt{16 + 1}=\sqrt{17}\)

Step5: <Use distance formula for \(RA\)>

\(RA=\sqrt{(2 - 1)^2+(2-(-2))^2}=\sqrt{(1)^2+(4)^2}=\sqrt{1+16}=\sqrt{17}\)

Answer:

\(PA=\sqrt{17}\), \(TP=\sqrt{17}\), \(TR=\sqrt{17}\), \(RA=\sqrt{17}\)