We know that:
∠ABC=∠BCA=xº
∠BDC=∠BCD=xº
1) we have to find the angle ∠DBC:
∠BDC+∠BCD+∠DBC=180º
xº+xº+∠DBC=180º
2xº+∠DBC=180º
∠DBC=180º-2xº
2) we have to find the angle ∠ABD:
∠ABD=∠ABC-∠DBC
∠ABD=xº-(180º-2x)
∠ABD=xº-180º+2xº
∠ABD=3xº-180º
Answer: ∠ABD=3xº-180º
x(-1,3) y(3,0) z(-1,-2)
xy
m = (3 + 1)/(0 - 3) = 4/-3 = -4/3
y - 3 = -4/3(x + 1)
y - 3 = -4/3x - 4/3
xy: y = -4/3x + 5/3
yz
m = (0 + 2)/(3 + 1) = 2/4 = 1/2
y = 1/2(x - 3)
yz: y = 1/2x - 3/2
xz
m = (3 + 2)/(-1 + 1) = 5/0 = undefined
y - 3 = undefined(x + 1)
xz: x = -1
xy: y = -4x/3 + 5/3
yz: y = x/2 - 3/2
xz: x = -1
Answer:
idk if this is right or not, but I got africa...
Answer:
x = 0
y = -4
negative 4 cuz the line goes down, meaning it's negative
