Question

using slope concept, prove (-2,1),(0,3),(2,1), and (0-1) are the vertices of a parallelogram
ANS FAST PLZ HELP​

Answer 1

Answer:

let ABCD is a parallelogram where AB||CD and AD||BC

Since in a parallelogram side AB=sideCD

and side AD=side BC

so AB=root [(-2-1)^2+(-1-0)^2]=root 10

and CD=root [(4-1)^2+(3-2)^2]=root 10

so here AB=CD

Again AD=root[(-2-1)^2+(-1-2)^2]=root 18

and BC=root [(1-4)^2+(0-3)^2)]=root 18

so AD=BC

hence ABCD is a parallelogram with the given vertices

Answer 2

Answer:

search-icon-image

Class 11

>>Applied Mathematics

>>Differentiation

>>Introduction

>>The points ( - 2, 1), (0, 3), (2, 1) and

Question

Bookmark

The points (−2,1),(0,3),(2,1) and (0,−1) are the vertices of a ________.

Medium

Solution

verified

Verified by Toppr

Correct option is A)

If the points are P(x

1

,y

1

) and Q(x

2

,y

2

), the slope of the line joining PQ is

x

2

−x

1

y

2

−y

2

For points, A(−2,1),B(0,3),C(2,1) and D(0,−1),

Slope of AB=

0+2

3−1

=1

Slope of BC=

2−0

1−3

=−1

Slope of CD=

0−2

−1−1

=1

Slope of DA=

0+2

−1−1

=−1

Clearly slope of AB= Slope of CD, hence, AB∥CD.

Clearly slope of BC= Slope of DA, hence, BC∥DA.

Since, the set of opposite sides are parallel to each other.

Thus, □ABCD are the vertices of a parallelogram.

Was this answer helpful ?