Respuesta :
Solution
Part 1
We have the following points given:
A (-1,3) B(3,3) C(5, -1) and D(1, -1)
The diagonals for this case are:
AC and BD
Then we can find the midpoint with the following points:
A (-1,3) C(5,-1)
[tex]x=\frac{x_1+x_2}{2}=\frac{-1+5}{2}=2[/tex][tex]y=\frac{y_1+y_2}{2}=\frac{3-1}{2}=1[/tex]Then the midpoint would be:
(2,1)
Part 2
For this case we have the following points:
P(-3,2) Q(2,6) R(3,3) S(x,y)
We need to satisfy the following:
slope PS= slope QR
And replacing we got:
[tex]\frac{y-2}{x+3}=\frac{3-6}{3-2}[/tex]Solving we have the following equation:
(y-2) = (x+3)*-3
y -2= -3x-9
y= -3x -7
Then we can create this other equation:
slope PQ = slope RS
replacing we got:
[tex]\frac{y-3}{x-3}=\frac{6-2}{2+3}[/tex]Solving we got:
5(y-3) = 4(x-3)
5y-15 =4x-12
5y -4x = 3
Replacing the first equation we got:
5(-3x-7) -4x= 3
-15x -35 -4x = 3
-19 x= 38
x= -2
y= -3*(-2)-7= 6-7 =-1
Then the answer is:
S= (-2,-1)
Part 3
We have the following points:
A= (-3,4)
B= (3,2)
C=(2,-1)
D=(-4,1)
We can find the distance between points like this:
[tex]d_{AB}=\sqrt[]{(3+3)^2+(2-4)^2}=\sqrt[]{40}[/tex][tex]d_{BC}=\sqrt[]{(2-3)^2+(-1-2)^2}=\sqrt[]{10}[/tex][tex]d_{CD}=\sqrt[]{(-4-2)^2+(1+1)^2}=\sqrt[]{40}[/tex][tex]d_{DA}=\sqrt[]{(-4+3)^2+(1-4)^2}=\sqrt[]{10}[/tex]Then AB= DA and BC= CD
Then the answer is:
Parallelogram
Rectangle
