Without using distance formula, show that the points A(1, –2), B(3, 6), C(5, 10) and D(3, 2) are the vertices of a parallelogram.

By using slope formula,

m = $\dfrac{y$2 - y1}{x2 - x1}

Given, points A(1, –2), B(3, 6)

Substituting values we get,

$m_{AB} = \dfrac{6 - (-2)}{3 - 1} = \dfrac{8}{2} = 4$

Given, points C(5, 10) and D(3, 2)

Substituting values we get,

$m_{DC} = \dfrac{10 - 2}{5 - 3} = \dfrac{8}{2} = 4$

Given, points B(3, 6), C(5, 10)

Substituting values we get,

$m_{BC} = \dfrac{10 - 6}{5 - 3} = \dfrac{4}{2} = 2$

Given, points A(1, –2), D(3, 2)

Substituting values we get,

$m_{AD} = \dfrac{2 - (-2)}{3 - 1} = \dfrac{4}{2} = 2$

m_AB = m_CD and m_BC = m_AD

∴ AB is parallel to CD and BC is parallel to AD

Since both pairs of opposite sides are parallel, the quadrilateral ABCD is a parallelogram.

Hence, proved that ABCD is a parallelogram.