Question

I found the slopes and graphed it I just need help finding the lengths.Quadrilateral OPQR can be described as:square, rectangle, a rhombus, a parellogram, a generic quadraliteral

Answer 1

Given:

Point O=(-3,-3)

Point P=(3,-7)

Point Q=(5,-4)

Point R=(-1,0)

To determine the slope, we use the formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

To get the slope of OP, we let :

x1=-3

y1=-3

x2=3

y2=-7

So,

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-7-(-3)}{3-(-3)}=\frac{-4}{6}=-\frac{2}{3}$

To get the length of OP, we use the distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

where:

d=distance

So,

$\begin{gathered} d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ =\sqrt{(3-(-3))^2+(-7-(-3))^2} \\ Simplify \\ d=\sqrt{52} \\ d=2\sqrt{13} \end{gathered}$

Hence, the length of OP is:

$2\sqrt{13}$

For PQ, we let:

x1=3

y1=-7

x2=5

y2=-4

So,

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{-4-(-7)}{5-3}=1\frac{1}{2}=\frac{3}{2}$$\begin{gathered} d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ d=\sqrt{2^2+3^2}=\sqrt{13} \end{gathered}$

For QR, we let:

x1=5

y1=-4

x2=-1

y2=0

So,

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{0-(-4)}{-1-5}=\frac{4}{-6}=-\frac{2}{3}$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-6)^2+(4)^2}=\sqrt{52}=2\sqrt{13}$

For RO, we let:

x1=-1

y1=0

x2=-3

y2=-3

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{-3-0}{-3-(-1)}=\frac{-3}{-2}=1\frac{1}{2}=\frac{3}{2}$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-2)^2+(-3)^2}=\sqrt{13}$

Therefore, the lengths are:

$\begin{gathered} OP=2\sqrt{13} \\ PQ=\sqrt{13} \\ QR=2\sqrt{13} \\ RO=\sqrt{13} \end{gathered}$

The given Quadrilateral OPQR is a rectangle since the opposite sides are equal and parallel to each other.