Question

verify that parallelogram abcd with vertices a(-5 -1) b(-9,6) c(-1,5) d(3,-2) is a rhombus by showing that it is a parallelogram with perpendicular diagonals

Answer 1

Answer: Showed.

Step-by-step explanation:  Given vertices of parallelogram abcd are a(-5,-1), b(-9,6), c(-1,5) and d(3,-2).

Since it is a parallelogram, so opposite sides must be equal. That is,

ab = cd,  bc = ad.

We are to show abcd is a rhombus by showing that the diagonals ac and bd are perpendicular to each other.

Now,

$ab=\sqrt{(-5+9)^2+(-1-6)^2}=\sqrt{16+49}=\sqrt{65},\\\\bc= \sqrt{(-9+1)^2+(6-5)^2}=\sqrt{64+1}=\sqrt{65}.$

So, ab = bc = cd = da, with proves that all the sides are equal.

Now, slope of diagonal ac will be

$S_{ac}=\dfrac{5+1}{-1+5}=\dfrac{3}{2},$

and the slope of bd will be

$S_{bd}=\dfrac{-2-6}{3+9}=-\dfrac{2}{3}.$

Therefore,

$S_{ac}\times S_{bd}=-1.$

This shows that the diagonals ac and bd are perpendicular.

Thus, the parallelogram abcd is a rhombus.