Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 14

MATH152 April 2016

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Question A 14
Find the area of the parallelogram with vertices at $(2,-2)$, $(3,1)$, $(5,6)$, and $(4,3)$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The area of a parallelogram is equal to the determinant of the 2×2 matrix formed by the column vectors representing two sides of the parallelogram.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let ${\mathbf {v}}_{1}=(4,3)-(2,-2)=(5,6)-(3,1)=(2,5)$ and ${\mathbf {v}}_{2}=(3,1)-(2,-2)=(5,6)-(4,3)=(1,3)$. These two vectors are two sides of the parallelogram. The area of a parallelogram is equal to the determinant of the 2×2 matrix formed by the column vectors representing component vectors: ${\text{Area of the parallelogram}}={\begin{vmatrix}{\mathbf {v}}_{1}&{\mathbf {v}}_{2}\end{vmatrix}}={\begin{vmatrix}2&1\\5&3\end{vmatrix}}=\color {blue}1.$