prove by section formula that the points(10,-6)(2,-6)(-4,-2)(4,-2) are vertices of parallelogram

Question

this is the question of class 10 from ML AGGARWAL book in the chapter distance and section formula and i am not understanding how i prove by section formula.

Answer 1

It helps to draw a picture. Let's call (10,-6) point C, (2,-6) point D, (-4,-2) point A and (4,-2) point B. Our parallelogram will be ABCD, as we'll see. The line CD contains the common y value -6, so CD is parallel to the x axis. The line AB contains the common y value -2, so it, too, is parallel to the x axis. Therefore, since AB and CD are both parallel to the x axis, they are parallel to one another. A pair of parallel sides. The slope of line AD is the difference between the y values divided by the difference between the x values: (-6-(-2))/(2-(-4))=(-6+2)/(2+4)=-4/6=-2/3. The slope of line BC is similarly (-6-(-2))/(10-4)=(-6+2)/6=-4/6=-2/3. The lines have the same slope so AD is parallel to BC. Therefore ABCD is a parallelogram.