We must prove that a set of points that forms a quadrilateral is a parallelogram or not. We will do it using the slope formula and the distance between the two points formula.
- The distance between two points is given by the following formula:
[tex]d(A,B)=\sqrt[]{(x_B-x_A)^2+(y_B-y_A)^2_{}}[/tex]- The slope m of a line a connect two points with coordinates (x_A,y_A) and (x_B,y_B) is the following:
[tex]m(AB)=\frac{y_B-y_A}{x_B-x_A}[/tex]Now, from the graph, we see that the supposed parallelogram has the following sides: AD, DC, CB and BA.
To prove that this quadrilateral is a parallelogram we must prove:
1) First, the opposite sides have equal length.
2) Secondly, opposite sides have equal slopes.
1) We compute the length of the sides.
[tex]\begin{gathered} AD=d(A,D)=\sqrt[]{(4+6)^2+(-1+4)^2}=\sqrt[]{109} \\ DC=d(D,C)=\sqrt[]{(5-4)^2+(7+1)^2}=\sqrt[]{65} \\ CB=d(C,B)=\sqrt[]{(-4-5)^2+(4-7)^2^{}}=\sqrt[]{90} \\ BA=d(B,A)=\sqrt[]{(-6+4)^2+(-4-4)^2}=\sqrt[]{68} \end{gathered}[/tex]We see that opposite sides are not equal:
[tex]\begin{gathered} AD\ne CB \\ DC\ne BA \end{gathered}[/tex]So we failed in proving the first requirement of the quadrilateral to be a parallelogram.
So we conclude that the set of points of the figure that forms a quadrilateral is not a parallelogram.
