Answer:
Option C is correct.
Explanation:
Rhombus states that a parallelogram with four equal sides and sometimes one with no right angle.
Given: The coordinate of the vertices of quadrilateral ABCD are A(−6, 3) , B(−1, 5) , C(3, 1) , and D(−2, −2) .
The condition for the segment [tex](x_{1},y_{1})[/tex], [tex](x_{2},y_{2})[/tex] to be parallel to [tex](x_{3},y_{3})[/tex], [tex](x_{4},y_{4})[/tex] is matching slopes;
[tex]\frac{y_{2}-y_{1}}{x_{2}-x_{1}}= \frac{y_{4}-y_{3}}{x_{4}-x_{3}}[/tex] or
[tex](y_{2}-y_{1}) \cdot (x_{4}-x_{3}) =(y_{4}-y_{3}) \cdot (x_{2}-x_{1})[/tex] ....[1]
So, we have to check that [tex]AB || CD[/tex] and [tex]AD || BC[/tex]
First check [tex]AB || CD[/tex]
A(−6, 3) , B(−1, 5) , C(3, 1) , and D(−2, −2)
substitute in [1],
[tex](5-3) \cdot (-2-3) = (-2-1) \cdot (-1-(-6))[/tex]
[tex]2 \cdot -5 = -3 \cdot 5[/tex]
-10 ≠ -15
Similarly,
check [tex]AD || BC[/tex]
A(−6, 3) , D(−2, −2) , B(−1, 5) and C(3, 1)
Substitute in [1], we have
[tex](-2-3) \cdot (3-(-1)) = (1-5) \cdot (-2-(-6))[/tex]
[tex]-5 \cdot 4 = -4 \cdot 4[/tex]
-20 ≠ -16.
Both pairs of sides are not parallel,
therefore, Quadrilateral ABCD is not a rhombus because there are no pairs of parallel sides.
