Question

The set of points (-3, 4), (-1, 1), (-3,-2), and (-5,1) identifies the vertices of a quadrilateral. Which is the most specific description to tell which figure the points form?
parallelogram
please help
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Answer 1

Answer:

Option (4). Rhombus

Step-by-step explanation:

From the figure attached,

Distance AB = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

                     = $\sqrt{(1-4)^2+(-5+3)^2}$

                     = $\sqrt{(-3)^2+(-2)^2}$

                     = $\sqrt{13}$

Distance BC = $\sqrt{(4-1)^2+(-3+1)^2}$

                     = $\sqrt{9+4}$

                     = $\sqrt{13}$

Distance CD = $\sqrt{(-2-1)^2+(-3+1)^2}$

                     = $\sqrt{9+4}$

                     = $\sqrt{13}$

Distance AD = $\sqrt{(1+2)^2+(-5+3)^2}$

                     = $\sqrt{9+4}$

                     = $\sqrt{13}$

Slope of AB ($m_{1}$) = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

                           = $\frac{4-1}{-3+5}$

                           = $\frac{3}{2}$

Slope of BC ($m_{2}$) = $\frac{4-1}{-3+1}$

                            = $-\frac{3}{2}$

If AB and BC are perpendicular then,

$m_{1}\times m_{2}=-1$

But it's not true.

[$m_{1}\times m_{2}=(\frac{3}{2})(-\frac{3}{2})$ = -$\frac{9}{4}$]

It shows that the consecutive sides of the quadrilateral are not perpendicular.

Therefore, ABCD is neither square nor a rectangle.

Slope of diagonal BD = $\frac{4+2}{-3+3}$

                                    = Not defined (parallel to y-axis)

Slope of diagonal AC = $\frac{1-1}{-1+5}$

                                    = 0 [parallel to x-axis]

Therefore, both the diagonals AC and BD will be perpendicular.

And the quadrilateral formed by the given points will be a rhombus.