Find the area of the parallelogram with vertices A(-3,0), B(-1,7),

Michelle Isakson 2021-11-16 Answered
Find the area of the parallelogram with vertices A(-3,0), B(-1,7), C(9,6), and D(7,-1)

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Expert Answer

Phisecome
Answered 2021-11-17 Author has 428 answers
Area of parallelogram with vertices
(-3,0), (-1,7), (9,6) and (7,-1) is given by
\(\displaystyle{A}={\frac{{{1}}}{{{2}}}}{\left[{x}_{{1}}{y}_{{2}}+{x}_{{2}}{y}_{{3}}+{x}_{{3}}{y}_{{4}}+{x}_{{4}}{y}_{{1}}-{x}_{{2}}{y}_{{1}}-{x}_{{3}}{y}_{{1}}-{x}_{{3}}{y}_{{2}}-{x}_{{4}}{y}_{{3}}-{x}_{{1}}{y}_{{4}}\right]}\)
Here, \(\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}={\left(-{3},{0}\right)}\)
\(\displaystyle{\left({x}_{{2}},{y}_{{2}}\right)}={\left(-{1},{7}\right)}\)
\(\displaystyle{\left({x}_{{3}},{y}_{{3}}\right)}={\left({9},{6}\right)}\)
\(\displaystyle{\left({x}_{{4}},{y}_{{4}}\right)}={\left({7},-{1}\right)}\)
\(\displaystyle{A}={\frac{{{1}}}{{{2}}}}{\mid}{\left(-{3}\right)}{\left({7}\right)}+{\left(-{1}\right)}{6}+{9}\times{\left(-{1}\right)}+{7}\times{\left({0}\right)}-{9}\times{7}-{7}\times{6}-{\left(-{3}\right)}{\left(-{1}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{\left|-{21}-{6}-{9}+{0}-{0}-{63}-{42}-{3}\right|}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{\left({144}\right)}={72}\)
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