Question

prove that the quadrilateral whose vertices are the points A(-1,1), B(-3,4), C(1,5) and D(3,2) is a parallelogram.

Answer 1

Answer:

Since $\overrightarrow{AB} = \overrightarrow{DC}$ and $\overrightarrow{AD} = \overrightarrow{BC}$, then the quadrilateral ABCD is a parallelogram.

Step-by-step explanation:

First, we label each point of the quadrilateral with the help of a graphing tool. If the quadrilateral ABCD is a parallelogram, then $\overrightarrow{AB} = \overrightarrow{DC}$ and $\overrightarrow{AD} = \overrightarrow{BC}$. If we know that $A(x,y) =(-1, 1)$, $B(x,y) =(-3,4)$, $C(x,y) = (1,5)$ and $D(x,y) = (3,2)$, then the measure of each vector is, respectively:

$\overrightarrow{AB} = (-3,4)-(-1,1)$

$\overrightarrow{AB} = (-2, 3)$

$\overrightarrow{DC} = (1,5)-(3,2)$

$\overrightarrow{DC} = ( -2,3)$

$\overrightarrow{AD} = (3,2)-(-1,1)$

$\overrightarrow{AD} = (4, 1)$

$\overrightarrow{BC} = (1,5)-(-3,4)$

$\overrightarrow{BC} = (4, 1)$

Since $\overrightarrow{AB} = \overrightarrow{DC}$ and $\overrightarrow{AD} = \overrightarrow{BC}$, then the quadrilateral ABCD is a parallelogram.