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The coordinates of the vertices of quadrilateral ABCD ar A(-6,-1), B(-5,2), C(0,-2), and D(-1,-5). Parker states that quadrilateral ABCD is a parallelogram. Prove or disprove Parker's statement.

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We are here given the coordinates of the vertices of a quadrilateral ABCD which are ,

  • A(-6,-1)
  • B(-5,2)
  • C(0,-2)
  • D(-1,-5)

And we need to prove or disprove that it is a llgm . We can do this using the distance formula. We can find the distance between adjacent points and if the measure of opposite sides comes equal then it is a ||gm .

Here opposite sides are AB and CD , and BC and AD .

Distance formula:-


\longrightarrow d =√((x_2-x_1)^2+(y_2-y_1)^2)

Distance between A and B :-


\longrightarrow d_(AB)= √( (-6+5)^2+(-1-2)^2)\\


\longrightarrow d_(AB)=√(-1^2+-3^2)\\


\longrightarrow d_(AB)= √(1+9)=\boxed{√(10)}

Distance between B and C :-


\longrightarrow d_(BC)=√((-5-0)^2+(2+2)^2)\\


\longrightarrow d_(BC)=√(-5^2+4^2)\\


\longrightarrow d_(BC)=√(25+16)=\boxed{√(41)} \\

Distance between C and D :-


\longrightarrow d_(AD)= √((0+1)^2(-2+5)^2)\\


\longrightarrow d_(AD)= √(1^2 +3^2)\\


\longrightarrow d_(AD)=√(1+9)=\boxed{√(10)} \\

Distance between A and D :-


\longrightarrow d_(AD)=√((-6+1)^2+(-5+1)^2)\\


\longrightarrow d_(AD)=√(-5^2+-4^2)\\


\longrightarrow d_(AD)=√(25+16)=\boxed{√(41)} \\

Since the opposite sides are equal, the given coordinates are of a llgm .

And we are done!

User Bakkot
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1 vote

Answer:

A parallelogram is a simple (non-self-intersecting) quadrilateral with two pairs of parallel sides. Plotting the points given above would give as a parallelogram where line made by AB is parallel and congruent with line CD. Hope this answers the question.

Explanation:

User Shalem
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