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Verified answer
A(3,4)B(2,2)
C(6,0)
garis AC --> (y-y1)/(y2-y1) = (x - x1)/(x2-x1)
(y-4)/(0-4) = (x - 3)/(6-3)
(y-4) / (-4) = (x -3) /(3)
3(y-4) = -4(x-3)
3y - 12 = -4x + 12
4x + 3y - 24 = 0
jarak B(2,2) ke AC = jarak B (2,2) ke 4x + 3y - 24=0
d = |ax + by + c/(√x²+y²)|
d = |4(2)+3(2) - 24/ √(2²+2²)|
d = |-10 /√8|
d = 10/8 √8 = (10/8)(2)√2 = 5/2 √2
2. L memotong AC sama besar
AL = LC
AL : LC = 1 : 1