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=(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=(y_(2)-y_(1))/(x_(2)-x_(1))=(-7-(-3))/(3-(-3))=(-4)/(6)=-(2)/(3)`

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

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

where:

d=distance

So,

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

Hence, the length of OP is:

`2√(13)`

For PQ, we let:

x1=3

y1=-7

x2=5

y2=-4

So,

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

For QR, we let:

x1=5

y1=-4

x2=-1

y2=0

So,

`m=(y_2-y_1)/(x_2-x_1)=(0-(-4))/(-1-5)=(4)/(-6)=-(2)/(3)`
`d=√((x_2-x_1)^2+(y_2-y_1)^2)=√((-6)^2+(4)^2)=√(52)=2√(13)`

For RO, we let:

x1=-1

y1=0

x2=-3

y2=-3

`m=(y_2-y_1)/(x_2-x_1)=(-3-0)/(-3-(-1))=(-3)/(-2)=1(1)/(2)=(3)/(2)`
`d=√((x_2-x_1)^2+(y_2-y_1)^2)=√((-2)^2+(-3)^2)=√(13)`

Therefore, the lengths are:

`\begin{gathered} OP=2√(13) \\ PQ=√(13) \\ QR=2√(13) \\ RO=√(13) \end{gathered}`

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