Question

Question: 11. Find the area of the parallelogram determined by the points \(\left( {1,4} \right),\)\(\left( { - 1,5} \right),\)\(\left( {3,9} \right),\) and \(\left( {5,8} \right)\). How can you tell that the quadrilateral determined by the points is actually a parallelogram?

Short answer

The area of the parallelogram is 12 square units. And the quadrilateral determined by the points is actually a parallelogram since one of the nonzero points can be written as the sum of the other two nonzero points.

Step-by-step solution

Translate the figure

First, translate one of the vertices to the origin. That is, subtract the vertex \(\left( {1,4} \right)\) from all four vertices \(\left( {1,4} \right),\left( { - 1,5} \right),\left( {3,9} \right),\) and \(\left( {5,8} \right)\). The new vertices so obtained are 0,\({v_1} = \left( { - 2,1} \right),\) \({v_2} = \left( {2,5} \right),\) and \({v_3} = \left( {4,4} \right)\).

Determine if the translated figure is a parallelogram

\(\begin{array}{c}{v_1} + {v_3} = \left( {\begin{array}{*{20}{c}}{ - 2}\\1\end{array}} \right) + \left( {\begin{array}{*{20}{c}}4\\4\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 2 + 4}\\{1 + 4}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}2\\5\end{array}} \right)\\{v_1} + {v_3} = {v_2}\end{array}\)

Note that the translated figure will be a parallelogram if and only if one of \({v_1},{v_2},\) and \({v_3}\) is the sum of the other two vectors.

Hence, this parallelogram determined by the columns of \(A = \left( {\begin{array}{*{20}{c}}{ - 2}&4\\1&4\end{array}} \right)\).

Find the area

\(\begin{array}{c}\left| {\det A} \right| = \left| {\det \left[ {\begin{array}{*{20}{c}}{ - 2}&4\\1&4\end{array}} \right]} \right|\\ = \left| { - 8 - 4} \right|\\ = \left| { - 12} \right|\\\left| {\det A} \right| = 12\end{array}\)

By Theorem 9, the area of the parallelogram is 12 square units.