Question

Prove that the points A(-2,-1), B(1,0), C(4,3) and D(1,2) are the vertices of a parallelogram ABCD.

Answer 1

Hint: Start by drawing the diagram, followed by proving DC is parallel to AB and AD is parallel to BC. If each pair of opposite sides of a quadrilateral is parallel, then the quadrilateral is a parallelogram. You must use the slopes of the edges of the parallelogram to show that they are parallel.

Complete step-by-step answer:
Let us start by drawing the diagram for a better visualisation of the situation given in the question.

Now to start with the question, we will try to prove that AD is parallel to BC and AB is parallel to DC. So, let us first find the slope of all the four edges.
We know, the slope of a line passing through points $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ .
Therefore, the sole of:
\[\begin{align}
  & {{m}_{AD}}=\dfrac{2-(-1)}{1-(-2)}=\dfrac{3}{3}=1 \\
 & {{m}_{AB}}=\dfrac{0-(-1)}{1-(-2)}=\dfrac{1}{3} \\
 & {{m}_{DC}}=\dfrac{3-2}{4-1}=\dfrac{1}{3} \\
 & {{m}_{BC}}=\dfrac{3-0}{4-1}=\dfrac{3}{3}=1 \\
\end{align}\]
Now, we know that two lines are parallel, if their slopes are equal. So, from the above results, we can see that the slope of BC is equal to the slope of AD. Also, the slope of AB is equal to the slope of DC. So, we can conclude that AD || BC and AB || DC. As we have shown that each pair of opposite sides of the quadrilateral ABCD is parallel, so the quadrilateral ABCD is a parallelogram.

Note: It is prescribed that you learn all the identities related to a straight line and its slopes, especially be careful with the formula of the slope as students generally get confused about the denominator and numerator of the formula. Also, learn the properties of parallelograms, including squares, as they might also be needed for solving such problems as we used in the above question.