" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
sejajar= gradien sama
gradien garis k=
y-y1/x-x1=14-8/4-2=6/2=3
(11,a) jadiin (x1,y1)nya
misal kita pake (x,y)nya (2,8)
y-y1/x-x1= 3
8-a/2-11 = 3
8-a/(-9) = 3
———— x (-9)
8-a=-27
-a=-27-8
a=35
d
19
tegak lurus: m1 x m2 = -1
m1 (gradien garis awal yg di soal)
3x-2y+6=0
rumus gradien:
-x/y= -3/-2 = 3/2
brarti 3/2 x m2 = -1
jadi m2 = -2/3
masukin rumus persamaan
y-y1=m(x-x1)
(x1,y1)nya dari (6,-2)
y-(-2)=-2/3(x-6)
y+2=-2/3(x-6)
—————— x 3
3y+6=-2(x-6)
3y+6 = -2x +12
3y+2x-6=0
b
20
3x-2y=16
2x+5y=-2
————+
5x+3y= 14
b
no 20 caranya banyak, ada eliminasi, subsitusi, dll. cari aja diinternet yaa, soalnya ga semua soal dg tipe kaya gini bisa pake cara yg diatas