Answer:
Step-by-step explanation:
Using the section formula, if a point (x,y) divides the line joining the points (x
1
,y
1
) and (x
2
,y
2
) in the ratio m:n, then
(x,y)=(
m+n
mx
2
+nx
1
,
m+n
my
2
+ny
1
)
The vertices of the triangle are given to be (x
1
,y
1
),(x
2
,y
2
) and (x
3
,y
3
). Let these vertices be A,B and C respectively.
Then the coordinates of the point P that divides AB in l:k will be
(
l+k
lx
2
+kx
1
,
l+k
ly
2
+ky
1
)
The coordinates of point which divides PC in m:k+l will be
⎩
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎧
m+k+l
mx
3
+(k+l)
(l+k)
lx
2
+kx
1
,
m+k+l
my
3
+(k+l)
(l+k)
ly
2
+ky
1
⎭
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎫
⇒(
m+k+l
kx
1
+lx
2
+mx
3
,
m+k+l
ky
1
+ly
2
+my
3
)
Distribute, then you get 3x+6-2x 4 then combine like terms and use your graphing calculator to graph
Answer:
Answer for b is BFC and CFD
Answer for c is BFD and BFA
Answer for d is BFD and AFE
I don't know about a sorry
Answer:
B. Yes, by the AA Similarity Postulate
Step-by-step explanation:
This triangles are both similar based on the AA Similarity Postulate. We know that all angles are equal because...
m∠ABE = m∠CBD = 42° (because they are veridical/opposite angels)
m∠AEB = 180° - m∠ABE - m∠EAB = 180° - 42° - 53° = 85°
Then from here we know that
m∠BDC = 180° - 85° - m∠CBD
m∠BDC = 95° - 42°
m∠BDC = 53°
From here we see that they are similar because of the AA Similarity Postulate, since
m∠ABE = m∠CBD = 42°
m∠BDC = m∠EAB = 53°
m∠AEB = m∠BCD = 85°