Answer: The area of the given parallelogram is 15 square units.
Step-by-step explanation:
Let ABCD is the parallelogram shown in the graph in which A≡(-2,-1), B≡(-3,3), C≡(1,2) and D≡(2,-2)
Since, AC is the diagonal of the parallelogram,
Thus, by the property of parallelogram,
Area of triangle ABC = Area of triangle ADC
Since, Area of parallelogram ABCD = Area of triangle ABC + Area of triangle ADC
= Area of triangle ABC + Area of triangle ABC
= 2 ( area of triangle ABC )
Since, the area of triangle ABC
[tex]=\frac{1}{2}|-2(3-2)-3(2-(-1))+1(-1-3)|[/tex]
[tex]=\frac{1}{2}|-2(1)-3(2+1)+1(-4)|[/tex]
[tex]=\frac{1}{2}|-2-3\times 3 -4|[/tex]
[tex]=\frac{1}{2}|-2-9-4|[/tex]
[tex]=\frac{1}{2}|-15|=\frac{15}{2}\text{ square unit}[/tex]
⇒ Area of parallelogram ABCD = [tex]2\times \frac{15}{2}[/tex]
[tex]=\frac{30}{2}[/tex]
[tex]=15\text{ square unit}[/tex]
Hence, The area of the given parallelogram is 15 square units.
