# Find the area of the parallelogram with vertices A(−3, 0),

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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limacarp4
Given vertices A(−3, 0), B(−1, 7), C(9, 6), D(7, −1)
$\stackrel{\to }{AB}=\left(-1-\left(-3\right),7-0\right)=\left(2,7\right)$
$\stackrel{\to }{BC}=\left(9-\left(-1\right),6-7\right)=\left(10,-1\right)$ The area of parallelogram with adjacent sides AB and BC is the length of their cross product $|\stackrel{\to }{AB}×\stackrel{\to }{BC}|$
Now
$|\stackrel{\to }{AB}×\stackrel{\to }{BC}|=|\begin{array}{ccc}i& j& k\\ 2& 7& 0\\ 10& -1& 0\end{array}|$
$=\left(0-0\right)i-\left(0-0\right)j+\left(-2-70\right)k$
$=-72k$
Therefore, the area is,
$|\stackrel{\to }{AB}×\stackrel{\to }{BC}|=\sqrt{{\left(-72\right)}^{2}}=72$
###### Not exactly what you’re looking for?
levurdondishav4
$=\frac{1}{2}\left[-3\left(7-6\right)-1\left(-6-63\right)\right]$
$=\frac{1}{2}\left[-3\left(1\right)-1\left(-66\right)\right]$
$=\frac{1}{2}\left[-3+66\right]$
$=\frac{1}{2}\left[63\right]$
$=\frac{63}{2}$ sq units
$=\frac{1}{2}\left[-3\left(6+1\right)-1\left(-9-42\right)\right]$
$=\frac{1}{2}\left[-3\left(7\right)-1\left(-51\right)\right]$
$=\frac{1}{2}\left[-21+51\right]$
$=\frac{1}{2}\left[30\right]$
$=12$ sq units
Area of parallelogram abcd = Area of triangle abc + Area of triangle acd
$=\frac{63}{2}+12$
$=\frac{63+24}{2}$
$=\frac{87}{2}$ sq units.
Therefore, the area of the parallelogram is $\frac{87}{2}$ sq. units.