Question

find the type of quadrilateral if points a(-4,-2),b(-3,-7),c(3,-2),d(2,3) are joined serially
​

Answer 1

Answer:

Given,

A ≡ (-4, -2) , B ≡ (-3, -7) , C ≡ (3, -2) and D ≡ (2,3)

First of all we have to find side length AB, BC , CD and DA

So, AB = \bf{\sqrt{(-3+4)^2+(-7+2)^2}=\sqrt{1^2+(-5)^2}=\sqrt{26}}

(−3+4)

2

+(−7+2)

2

=

1

2

+(−5)

2

=

26

BC = \bf{\sqrt{(3+3)^2+(-2+7)^2}=\sqrt{6^2+5^2}=\sqrt{61}}

(3+3)

2

+(−2+7)

2

=

6

2

+5

2

=

61

CD = \bf{\sqrt{(2-3)^2+(3+2)^2}=\sqrt{(-1)^2+5^2}=\sqrt{26}}

(2−3)

2

+(3+2)

2

=

(−1)

2

+5

2

=

26

DA=\bf{\sqrt{(2+4)^2+(3+2)^2}=\sqrt{6^2+5^2}=\sqrt{61}

Here you see, AB = CD and BC = DA

For more information require to identify ,

Find midpoint of AC and BD

Midpoint of AC ={(-4 +3)/2, (-2 - 2)/2 } = (-1/2 ,-2)

Midpoint of BD = {(-3 + 2)/2 , (-7 + 3)/2} = (-1/2, -2)

E.,g., Midpoint of AC = midpoint of BD

It is clear that ABCD is parallelogram .

Now, find AC for identifying it is rectangle or not .

AC = \bf{\sqrt{(3+4)^2+(-2+2)^2}=7}

(3+4)

2

+(−2+2)

2

=7

We see, AB, AC and BC doesn't follow Pythagoras theorem,

So, ABCD doesn't rectangle .

Hence, ABCD is parallelogram

Answer 2

Answer:

the answer is to long so i am text the concept i hope u understand

Step-by-step explanation:

let first of all ab and make their corrdinates is like x1 x2 y1 y2 =√(x2-x1) square +(y2-y1) square

this is a formula revise it and then like bc, cd, da, db you get your answer