Answer:
Step-by-step explanation:
11)
find the slope use m1= y2-y1 / x2-x1 where p1=(2,7) p2=(6,12)
m=12-7 / 6-2 = 5/4
p3 = (13,130 p4 (9,8)
m2= 8-13 / 9-13 = -5 / -4 = 5/4
m1 =m2 so this is a parallelogram
(attached Bdrawing1)
12)
find slope and distance
p1=(-5,-6) p2=(5,2) p3=(4,-4) p4=(-6,-12)
m1=2-(-6) / 5-(-5) = 2+6 / 5+5 = 8/10 = 4/5
m2 = -12-(-4) / -6-4 = -12 +4 / -10 = -8 / -10 = 4/5
yes this looks the same in slop... check distance now
dist.= [tex]\sqrt{(x2-x1)^{2} + (y2-y1)^{2} }[/tex]
dist1 =[tex]\sqrt{(5-(-5))^{2}+(2-(-6))^{2} }[/tex]
dist1=[tex]\sqrt{10^{2}+8^{2} }[/tex]
dist1 = [tex]\sqrt{164}[/tex]
dist1= 12.806...
dist2 = [tex]\sqrt{(-6-4)^{2}+(-12-(-4))^{2} }[/tex]
dist2 = [tex]\sqrt{(-10)^{2}+(-8)^{2} }[/tex]
dist2 = [tex]\sqrt{164}[/tex]
dist2= 12.806...
yes this is also a parallele a gram..
(attachced Bdrawing2)
