Answer:
A-7
Step-by-step explanation:
Count, Add, then sum it up or just take my answer.
The area of the triangle QRS vertices have coordinates as (-1,2), (1,-4) and (-2,-2) is given by: Option: 7 sq. units.
Suppose the vertices of the considered triangle ABC are on [tex]A(A_x,A_y), B(B_x,B_y), C(C_x,C_y)[/tex], then, the area of the triangle is given by:
[tex]Area = \dfrac{|A_x(B_y-C_y) + B_x(C_y-A_y) + C_x(A_y-B_y)|}{2}[/tex]
For this case, we're specified that:
Coordinates of the vertices of the triangle QRS are:
(-1,2), (1,-4) and (-2,-2)
Thus, its area is evaluated as:
[tex]Area = \dfrac{|-1(-4-(-2)) + 1(-2-2) + (-2)(2-(-4))|}{2} = \dfrac{|2-4-12|}{2}\\\\Area = \dfrac{14}{2} = 7 \: \rm unit^2[/tex]
Thus, the area of the triangle QRS vertices have coordinates as (-1,2), (1,-4) and (-2,-2) is given by: Option: 7 sq. units.
Learn more about area of a triangle from the coordinates of its vertices here:
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