Question

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.

Answer 1

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!

Answer 2

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: