Question

select the correct graph of parallelogram abcd on a coordinate plane with vertices at a(3,4), b(3,2), c(-1,-2), and d(-3,0).

Explanation:

Step1: Recall the properties of a parallelogram

In a parallelogram \(ABCD\), \(AB\parallel CD\) and \(AB = CD\), \(AD\parallel BC\) and \(AD=BC\). We can also use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) to check the lengths of the sides.
For two points \((x_1,y_1)\) and \((x_2,y_2)\), the slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\) to check the parallelism (equal slopes for parallel lines).
Let's calculate the lengths:

• Length of \(AB\): Given \(A(3,4)\) and \(B(5,2)\), \(d_{AB}=\sqrt{(5 - 3)^2+(2 - 4)^2}=\sqrt{4 + 4}=\sqrt{8}=2\sqrt{2}\)
• Length of \(CD\): Given \(C(-1,-2)\) and \(D(-3,0)\), \(d_{CD}=\sqrt{(-3+1)^2+(0 + 2)^2}=\sqrt{4 + 4}=\sqrt{8}=2\sqrt{2}\)
• Length of \(AD\): Given \(A(3,4)\) and \(D(-3,0)\), \(d_{AD}=\sqrt{(-3 - 3)^2+(0 - 4)^2}=\sqrt{36+16}=\sqrt{52} = 2\sqrt{13}\)
• Length of \(BC\): Given \(B(5,2)\) and \(C(-1,-2)\), \(d_{BC}=\sqrt{(-1 - 5)^2+(-2 - 2)^2}=\sqrt{36 + 16}=\sqrt{52}=2\sqrt{13}\)

Step2: Check the slopes

• Slope of \(AB\): \(m_{AB}=\frac{2 - 4}{5 - 3}=\frac{-2}{2}=-1\)
• Slope of \(CD\): \(m_{CD}=\frac{0+2}{-3 + 1}=\frac{2}{-2}=-1\)
• Slope of \(AD\): \(m_{AD}=\frac{0 - 4}{-3 - 3}=\frac{-4}{-6}=\frac{2}{3}\)
• Slope of \(BC\): \(m_{BC}=\frac{-2 - 2}{-1 - 5}=\frac{-4}{-6}=\frac{2}{3}\)

Since \(AB\parallel CD\) ( \(m_{AB}=m_{CD}=-1\) ) and \(AD\parallel BC\) ( \(m_{AD}=m_{BC}=\frac{2}{3}\) ) and \(AB = CD\), \(AD=BC\), we can also check the position of the points on the coordinate - plane.
Plot the points \(A(3,4)\), \(B(5,2)\), \(C(-1,-2)\), \(D(-3,0)\) on the coordinate - plane.
The correct graph is the one where the order of the vertices \(A\), \(B\), \(C\), \(D\) forms a parallelogram with the calculated side - length and slope properties.

Answer:

C.