Question

prove that the points (-7, -3), (5, 10),(15, 8) and (3, -5) taken in order are the corners of a parallelogram. find it's area​

Answer 1

Answer:

First you follow me please........

Answer 2

Answer:

120 Square units

Step-by-step explanation:

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

Using the distance formula d = $\sqrt{(x2 - x1)^{2} + (y2 - y1)^{2}}$

$AB = \sqrt{(5+7)^{2} +(10+3)^{2}} = \sqrt{12^{2} + 13^{2}} = \sqrt{144+169} = \sqrt{313}$

$BC = \sqrt{(15-5)^{2} +(8-10)^{2}} = \sqrt{10^{2} + (-2)^{2}} = \sqrt{100 + 4} = \sqrt{104}$

$CD = \sqrt{(3-15)^{2} +(-5-8)^{2}} = \sqrt{(-12)^{2} + (-13)^{2}} = \sqrt{144+169} = \sqrt{313}$$AD = \sqrt{(3+7)^{2} +(-5+3)^{2}} = \sqrt{10^{2} + (-2)^{2}} = \sqrt{100 + 4} = \sqrt{104}$

So, AB = CD =  $\sqrt{313}$  and BC = DA =  $\sqrt{104}$

​i.e., The opposite sides are equal. Hence ABCD is a parallelogram.

Area of Parallelogram = B x H

Let us consider the positive values of C(15,8) as base and height.

Area = 15 x 8 = 120 Square units