Question

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

Answer 1

If ABCD is a parallelogram,  AB must be parallel to CD and AD must be parallel to BC.

Two lines are parallel only when they have same slope.

   If AB and CD are parallel, their slope must equal.

          Using,  slope = (y₂ - y₁)/(x₂ - x₁)

Slope(AB) = (0 - 1)/(-1 - 3)

                 = 1/4

Slope(CD) = (1 - 2)/(-2 - 2)

                 = 1/4

As slope of AB and CD are equal, AB and CD are parallel.

Similarly,

   If AD and BC are parallel, their slope must equal.

          Using,  slope = (y₂ - y₁)/(x₂ - x₁)

Slope(AD) = (0 - 1)/(-1 - (-2))

                 = -1

Slope(BC) = (2 - 1)/(2 - 3)

                 = -1

As slope of AD and BC are equal, AD and BC are parallel.

         Hence the given quadrilateral is a parallelogram. (being AB || CD  &  AD || BC)

Therefore, A, B, C and D are the vertices of a parallelogram.

Answer 2

Answer:

$\qquad\qquad\underline{\textsf{\textbf{ \color{magenta}{Question :-} }}}$

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

$\qquad\qquad\underline{\textsf{\textbf{ \color{magenta}{Solution :-} }}}$

Let A(- 1, 0), B(3, 1), C(2, 2) and D(x, y) be the vertices of a parallelogram ABCD taken in order.

Since, the diagonals of a parallelogram bisect each other.

So, coordinate of the mid point of AC = coordinate of mid point of BD

⇒ [(-1 + 2)/2, (0 + 2)/2] = [(3 + x)/2, (y + 1)/2]

⇒ (1/2, 1) = [(3 + x)/2, (y + 1)/2]

(3 + x)/2 = ½ ⇒ x = - 2

Also (y + 1)/2 = 1 ⇒ y + 1 = 2

⇒ y = 1

The fourth vertex of parallelogram = (- 2, 1).