Question

Show that the points A (1, 0), B (5, 3), C (2, 7) and D (-2, 4) are the vertices of a parallelogram.

Answer 1

Let given points be A (1, 0), B (5, 3), C (2, 7) and D (-2, 4) and let the intersection of diagonals be E(x_m , y_m )

By midpoint formula.

x = \(\frac{x_1+x_2}2\), y = \(\frac{y_1+y_2}2\)

For midpoint of diagonal AC,

 x₁ = \(\frac{1 + 2}2\), y₁ = \(\frac{0+7}2\)

∴x₁ = \(\frac{3}2\), y₁ = \(\frac{7}2\)

midpoint of diagonal AC is (x₁, y₁) ≡ ( \(\frac{3}2\), \(\frac{7}2\) ) …(1) 

For midpoint of diagonal BD,

x₂ = \(\frac{5 - 2}2\), y₂ = \(\frac{3+4}2\)

∴x₂ = \(\frac{3}2\), y₂ = \(\frac{7}2\)

∴ midpoint of diagonal BD is (x_2, y₂ ) ≡ (\(\frac{3}2\), \(\frac{7}2\))…..(2)

Here, from 1 and 2 we say that midpoint of both the diagonals intersect at same point, ie ( \(\frac{3}2\),\(\frac{7}2\) ) 

But our intersection of diagonals is at E, which means that midpoint of diagonals intersect at single point, ie E(\(\frac{3}2\) , \(\frac{7}2\)) 

We know that if midpoints of diagonals intersect at single point, then quadrilateral formed by joining the points is parallelogram. 

Hence, 

our □ABCD is parallelogram.