Question

Q1. Show that the

points A (0, 0), B(3, 0),

C(4, 1) and D(1, 1)

form a parallelogram​

Answer 1

Step-by-step explanation:

Given points are

A(−2,−1),B(1,0),C(4,3) and D(1,2)

A quadrilateral is a parallelogram if the opposite sides are equal.

∴AB

2

=(−2−1)

2

+(−1−0)

2

=9+1

=10

BC

2

=(4−1)

2

+(3−0)

2

=9+9

=18

CD

2

=(1−4)

2

+(2−3)

2

9+1

=10

AD

2

=(−2−1)

2

+(−1−2)

2

=9+9

=18

∴AB=CD=

10

and BC=AD=

18

∴ The opposite sides of the quadrilateral ABCD are equal, the four points from a parallelogram

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Answer 2

ATQ, We have a quadrilateral ABCD with the coordinates A(0, 0), B(3, 0), C(4, 1) and D(1, 1).

For a quadrilateral to be a parallelogram, the opposite sides need to be equal. Therefore, let's try to find the distances of AB, BC, CD and AD.

Finding the distance of AB:

$\rm \Longrightarrow AB = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$\rm \Longrightarrow AB = \sqrt{(0 - 3)^2 + (0 - 0)^2}$

$\rm \Longrightarrow AB = \sqrt{(-3)^2}$

$\rm \Longrightarrow AB = \sqrt{9}$

$\rm \Longrightarrow AB = 3 \ sq.units$

Finding the distance of BC:

$\rm \Longrightarrow BC = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$\rm \Longrightarrow BC = \sqrt{(3 - 4)^2 + (0 - 1)^2}$

$\rm \Longrightarrow BC = \sqrt{(-1)^2 + (-1)^2}$

$\rm \Longrightarrow BC = \sqrt{1 + 1}$

$\rm \Longrightarrow BC = \sqrt{2} \ sq. units$

Finding the distance of CD:

$\rm \Longrightarrow CD = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$\rm \Longrightarrow CD = \sqrt{(4 - 1)^2 + (1 - 1)^2}$

$\rm \Longrightarrow CD = \sqrt{(3)^2 + (0)^2}$

$\rm \Longrightarrow CD = \sqrt{9}$

$\rm \Longrightarrow CD = 3 \ sq.units$

Finding the distance of DA:

$\rm \Longrightarrow DA = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$\rm \Longrightarrow DA = \sqrt{(1 - 0)^2 + (1 - 0)^2}$

$\rm \Longrightarrow DA = \sqrt{(1)^2 + (1)^2}$

$\rm \Longrightarrow DA = \sqrt{2} \ sq.units$

Here, AB = CD and BD = AD.

∴ ABCD is a parallelogram.