Question

Prove that the points (4, 5), (7, 6), (6, 3), (3, 2) are the vertices of a parallelogram. Is it a rectangle.

Answer 1

Let given points be A(4, 5), B(7, 6), C(6, 3), D(3, 2) 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 = \(\frac{4+6}2\) , y₁ = \(\frac{5+3}2\)

∴  x₁ = \(\frac{10}2\) = 5, y₁ = \(\frac{8}2\) = 4

∴ midpoint of diagonal AC is (x₁, y₁ ) ≡ (5, 4) …(1) 

For midpoint of diagonal BD,

x₂ = \(\frac{7+3}2\) , y₂ = \(\frac{6+2}2\)

∴  x₂ = \(\frac{10}2\) = 5, y₂ = \(\frac{8}2\) = 4

∴ midpoint of diagonal BD is (x₂, y₂ ) ≡ (5, 4) …(2) 

Here, 

from 1 and 2 we say that midpoint of both the diagonals intersect at same point, ie (5, 4) 

But our intersection of diagonals is at E, which means that midpoint of diagonals intersect at single point, ie E(5, 4) 

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. 

Now, 

we shall check whether □ABCD is rectangle. 

If the lengths of diagonals are same, then given quadrilateral is rectangle. 

By distance formula,

XY = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

For diagonal AC,

AC = \(\sqrt{(6-4)^2+(3-5)^2}\)

= \(\sqrt{4+4}\)

= \(2\sqrt2\) units

For diagonal BD,

AC = \(\sqrt{(7-3)^2+(6-2)^2}\)

 = \(\sqrt{16+16}\)

= \(4\sqrt2\) units

Here, AC ≠ BD, hence □ABCD is not rectangle.