By using the concept of slope, show that the points (– 2, – 1), (4

1.	By using the concept of slope, show that the points (– 2, – 1), (4, 0), (3, 3) and (– 3, 2) vertices of a parallelogram.
Answer» 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_CDand m_BC = m_AD i.e, AB\|\|CD and BC\|\|AD We know, If opposite side of a quadrilateral are parallel that it is a parallelogram. Hence, ABCD is a Parallelogram.

By using the concept of slope, show that the points (– 2, – 1), (4, 0), (3, 3) and (– 3, 2) vertices of a parallelogram.

Answer»

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_CDand m_BC = m_AD

i.e, AB||CD and BC||AD

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

Hence, ABCD is a Parallelogram.