Question

Without using the distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (– 3, 2) are the vertices of a parallelogram.

Answer 1

To Prove: Given points are of Parallelogram. 

Explanation: Let us Assume that we have points, A (– 2, – 1), B(4, 0), C(3, 3) and D(– 3, 2), are joining the sides as AB, BC, CD, and AD. 

The formula used: The slope of the line, m = \(\frac{y_2-y_1}{x_2-x_1}\)

Now, The slope of Line AB, m_AB = \(\frac{0-(-1)}{4-(-2)}\)

m_AB = \(\frac{1}{6}\) 

The slope of BC, m_BC =  \(\frac{3-0}{3-4}\) 

m_BC =  \(\frac{3}{-1}\) 

Now, The slope of Line CD, m_CD =  \(\frac{2-3}{-3-3}\) 

m_CD =  \(\frac{1}{6}\) 

The slope of AD, m_AD =  \(\frac{2-(-1)}{-3-(-2)}\) 

m_AD =  \(\frac{3}{-1}\) 

Here, We can see that, m_AB = m_CD and m_BC = m_AD 

i.e, AB||CD and BC||AD 

We know, If opposite side of a quadrilateral are parallel that it is parallelogram. 

Hence, ABCD is a Parallelogram.