Find the area of the parallelogram with vertices A(-3, 0), B(-1 , 3), C(5, 2), and D(3, -1).

Question

casincal · Accepted Answer

Calculations:We can choose any three of the supplied points and dind a two vector that expresses the sides of the parallelogram &quot;or one side and a diagonal of the parallelogram,&quot; we can choose the points B, C, and D.Furthermore, we can choose point B as the common point for the two sides &quot;the initial point of the two vectors,&quot; so we must find vector BD and vector BC using the equation below.BC=&amp;lt;5+1,2−3&amp;gt;=&amp;lt;6,−1&amp;gt;And, the other vector side BD isBD=&amp;lt;3+1,−1−3&amp;gt;=&amp;lt;4,−4&amp;gt;Furthermore, the cross product of both vectors isBC×BD=&amp;lt;6,−1&amp;gt;×&amp;lt;4,−4&amp;gt;=|ijk6−104−40|=i|−1040|−j|6040|+k|6−14−4|=0i−0j+k((6)(−4)−(1)(−4))=−20kKnowing the cross product of the parallelogram&#039;s two vectors, we may apply an equation to get the area.Area=|−20k|=20As a result, the parallelogram&#039;s area is 20 units squared.

xleb123 · Accepted Answer

Step 1:Given:Area=|12((x2−x1)(y4−y1)−(x4−x1)(y2−y1))|Let&#039;s calculate the area using the given vertices A(-3, 0), B(-1, 3), C(5, 2), and D(3, -1).We can label the coordinates as follows:A(−3,0), B(−1,3), C(5,2), and D(3,−1).Step 2:Now, we substitute these values into the formula:Area=|12((−1−(−3))(2−0)−(3−(−3))(3−0))|Simplifying further:Area=|12((2)(2)−(6)(3))|Area=|12(4−18)|Area=|12(−14)|Taking the absolute value:Area=12·14Finally, evaluating the expression:Area=7Therefore, the area of the parallelogram with vertices A(-3, 0), B(-1, 3), C(5, 2), and D(3, -1) is 7 square units.

Andre BalkonE · Accepted Answer

fudzisako · Accepted Answer

To find the area of the parallelogram with vertices A(-3, 0), B(-1, 3), C(5, 2), and D(3, -1), we can use the formula:

Area = | \frac{1}{2} \cdot (x_{1} y_{2} + x_{2} y_{3} + x_{3} y_{4} + x_{4} y_{1} - x_{2} y_{1} - x_{3} y_{2} - x_{4} y_{3} - x_{1} y_{4}) |

where

(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}), (x_{4}, y_{4})

are the coordinates of the four vertices of the parallelogram.
Substituting the given coordinates into the formula, we have:

Area = | \frac{1}{2} \cdot ((- 3) (3) + (- 1) (2) + (5) (- 1) + (3) (0) - (- 1) (0) - (5) (3) - (3) (- 1) - (- 3) (2)) |

Simplifying the expression inside the absolute value:

Area = | \frac{1}{2} \cdot (- 9 - 2 - 5 + 0 + 0 - 15 + 3 + 6) |

Area = | \frac{1}{2} \cdot (- 22) |

Area = | - 11 |

Therefore, the area of the parallelogram is

11

square units.

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

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