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1a. y+1=1/2(x-0) => y+1=1/2x => y=1/2x-1
1b. y-3=-1(x-0) => y=-x+3
2a. y-9=3/5(x-5) => y=3/5x-3+9 => y=3/5x+6
2b. y-3=-1/2(x-6) => y=-1/2x+3+3 => y=-1/2x+6
Rumus untuk soal 3 => (Y-Y1) /Y2-Y1 =(X-X1) /X2-X1
3a. Y+4/6+4=X+1/2+1 => Y+4/10=X+1/3
=> 3(Y+4)=10(X+1) => 3Y+12=10X+10 => 3Y=10X-2
=> Y=10/3x-2/3
3b. Y+5/3+5=X-8/1-8 => Y+5/8=X-8/-7
=> -7(Y+5)=8(X-8) => - 7Y-35=8X-64 => - 7Y=8X-29
=> Y=-8/7X-29/7