Question

Find the type of the quadrilateral if points A(-4,-2), B(-3,-7) C(3,-2) and D(2,3) are joined serially.

Answer 1

$\bf{Given,}$
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 = $\bf{\sqrt{(-3+4)^2+(-7+2)^2}=\sqrt{1^2+(-5)^2}=\sqrt{26}}$
BC = $\bf{\sqrt{(3+3)^2+(-2+7)^2}=\sqrt{6^2+5^2}=\sqrt{61}}$
CD = $\bf{\sqrt{(2-3)^2+(3+2)^2}=\sqrt{(-1)^2+5^2}=\sqrt{26}}$
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 = $\bf{\sqrt{(3+4)^2+(-2+2)^2}=7}$
We see, AB, AC and BC doesn't follow Pythagoras theorem,
So, ABCD doesn't rectangle .

Hence, ABCD is parallelogram

Answer 2

Answer:

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