It is the AAS Postulate.
Explanation: You’re given two right angles (because perpendicular sides make right angles).
This also means that the angle next to them are also right angles, because they are linear pairs (180-90=90)
The right angles are:
This is your first A.
Both triangles also share a similar angle:
This is your second A.
You’re given side BD is congruent to EC.
This is your S.
It’s AAS and not ASA because the order refers to how they are connected: they are congruent in the order that the first set of congruent angles connect to the next, then to the congruent side.
Y = -6x + 3
In y = mx + b form, the slope will be in ur m position...
y = mx + b
y = -6x + 3
so ur slope (the number in the m position) is -6
What we know:
line P endpoints (4,1) and (2,-5) (made up a line name for the this line)
perpendicular lines' slope are opposite in sign and reciprocals of each other
slope=m=(y2-y1)/(x2-x1)
slope intercept for is y=mx+b
What we need to find:
line Q (made this name up for this line) , a perpendicular bisector of the line p with given endpoints of (4,1) and (2,-5)
find slope of line P using (4,1) and (2,-5)
m=(-5-1)/(2-4)=-6/-2=3
Line P has a slope of 3 that means Line Q has a slope of -1/3.
Now, since we are looking for a perpendicular bisector, I need to find the midpoint of line P to use to create line Q. I will use the midpoint formula using line P's endpoints (4,1) and (2,-5).
midpoint formula: [(x1+x2)/2, (y1+y2)/2)]
midpoint=[(4+2)/2, (1+-5)/2]
=[6/2, -4/2]
=(3, -2)
y=mx=b when m=-1/3 slope of line Q and using point (3,-2) the midpoint of line P where line Q will be a perpendicular bisector
(-2)=-1/3(3)+b substitution
-2=-1+b simplified
-2+1=-1+1+b additive inverse
-1=b
Finally, we will use m=-1/3 slope of line Q and y-intercept=b=-1 of line Q
y=-1/3x-1
Answer:
its B ITHINK
Step-by-step explanation:
