dany jest trojkat ABC gdzie A(-2,3) B(-4,-1) C(2,0)
a)wyznacz rownie prostej AB
b)oblicz dlugosc wysokosci opuszczonej na bok AB
c)oblicz pole trojkata ABC
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A(-2;3)
B(-4;-1)
C(2;0)
a]
y=ax+b
3=-2a+b
-1=-4a+b
b=2a+3
-1=-4a+2a+3
2a=3+1
2a=4
a=2
b=2×2+3=7
równanie prostej AB:
y=2x+7
b]
y=2x+7
2x-y+7=0
A=2
B=-1
C=7
C(2;0)
h=I2×2+-1×0+7I/√2²+(-1)²=11/√5=11√5/5
a=√(-4+2)²+(-1-3)²=√4+16=√20=2√5
c]
pole=½ah=½×2√5×11√5/5=11j.²