Question

show that the following points taken in order from the veritices of the parallelogarm A(-7,-3) B(5,10) C(15,8) D(3,-5)​

Answer 1

Answer:

Hi,

Let the points A(-7,-3), B(5,10),

C(15,8) and D(3,-5) are vertices of a parallelogram.

We know that,

The diagonals of a parallelogram bisects each other.

So, the midpoint of diagonal AC

and DB should be same.

Now, we find mid points of AC and

DC by using (x1 + x2/2, y1 + y2/2)

formula

i) midpoint of AC = (-7+15/2, -3+8/2)

= (8/25/2)

= ( 4, 5/2) -----(1)

ii) mid point of DB = (3+5/2, -5+10/2)

= ( 8/2,5/2)

= ( 4, 5/2) -----(2)

Hence, midpoint of AC and mid point

DB is same.

Therefore,

the points A, B, C, D are vertices of

a parallelogram.

I hope this helps you.

:)

Answer 2

Answer:

Given :-

• The vertices of the parallelogram A(- 7, - 3) , B(5, 10) , C(15, 8) , D(3, - 5).

Show That :-

• The following points are parallelogram.

Formula Used :-

$\bigstar$ Distance Formula :-

$\mapsto \sf\boxed{\bold{\pink{d =\: \sqrt{\bigg(x_2 - x_1\bigg)^2 + \bigg(y_2 - y_1\bigg)^2}}}}$

Solution :-

Let, the points (- 7, - 3) , (5, 10), (15, 8) (3, - 5) represent the points A, B, C, D.

✭ In case of AB :-

$\implies \sf AB =\: \sqrt{(5 + 7)^2 + (10 + 3)^2}$

$\implies \sf AB =\: \sqrt{(12)^2 + (13)^2}$

$\implies \sf AB =\: \sqrt{(12 \times 12) + (13 \times 13)}$

$\implies \sf AB =\: \sqrt{144 + 169}$

$\implies \sf AB =\: \sqrt{313}$

✭ In case of BC :

$\implies \sf BC =\: \sqrt{(15 - 5)^2 + (8 - 10)^2}$

$\implies \sf BC =\: \sqrt{(10)^2 + (- 2)^2}$

$\implies \sf BC =\: \sqrt{(10 \times 10)+ \{(- 2) \times (- 2)\}}$

$\implies \sf BC =\: \sqrt{100 + 4}$

$\implies \sf BC =\: \sqrt{104}$

✭ In case of CD :

$\implies \sf CD =\: \sqrt{(3 - 15)^2 + (- 5 - 8)^2}$

$\implies \sf CD =\: \sqrt{(- 12)^2 + (- 13)^2}$

$\implies \sf CD =\: \sqrt{\{(- 12) \times (- 12)\} + \{(- 13) \times (- 13)\}}$

$\implies \sf CD =\: \sqrt{144 + 169}$

$\implies \sf CD =\: \sqrt{313}$

✭ In case of AD :

$\implies \sf AD =\: \sqrt{(3 + 7)^2 + (- 5 + 3)^2}$

$\implies \sf AD =\: \sqrt{(10)^2 + (- 2)^2}$

$\implies \sf AD =\: \sqrt{(10 \times 10) + \{(- 2) \times (- 2)\}}$

$\implies \sf AD =\: \sqrt{100 + 4}$

$\implies \sf AD =\: \sqrt{104}$

Here, we noticed that :

➲ AB = CD = √313

➲ BC = AD = √104

Hence opposite sides are equal.

∴ ABCD is a Parallelogram.