Question

Show that A (-4,-7), B (-1,2), C (8,5) and D (5,-4) are the vertices of a parallelogram.please give this answer step by step. ​

Answer 1

we know : Parallelogram property is opposite sides are equal .And diagonal bisect each other

Step-by-step explanation:

We know distance formula √(x2-x1)^2+(Y2-Y1)^2

For sides

Let, A(X1 , Y1) =A(-4,-7) and B (X2 ,Y2) =(-1,2)

AB = distance formula

AB=✓(-1+4)^2+ (2+7)^2

AB = ✓90 (after solving)

Let now B(X1,Y1) and C(X2, Y2)

BC =✓(8+1)^2 +(5-2)^2)

BC = ✓90

Let now C(X2,Y2) and D (X1,Y1)

CD= ✓(8-5)^2+(5+4)^2

CD=✓90

Let now D(X2,Y2) and A (X1,Y1)

DA=✓(5+4)^2 +(-4+7)^2

DA=✓90

Now for Diagonal

1. Let, A (X1,Y1) and C(X2 ,Y2)

AC=✓(8+4)^2 +(5+7)^2

AC=✓288

2. Let B(X1,Y1 ) and D (X2 ,Y2 )

BD=✓(5+1)^2+(-4-2)^2

BD=✓72

From above we can conclude all sides are equal but diagonal are unequal .Hence it is a rhombus and we know every rhombus is a parallelogram but vise versa is not true.

Therefore it is a Parallelogram. ---- proved