persamaan garis lurus y=x+2 didilatasi oleh [P(2,1),D(3)] bayangan garis lurus k memotong sumbu x di titik berabsis..

y = x + 2
P(a,b) = (2,1)
k = 3

x' = k(x -a) + b→ x' = 3(x-2) + 1
x' = 3x - 5 → 3x = x' + 5  → x = 1/3 x' + 5/3

y' = k(y-b) + b → y' = 3(y-1) + 1
y' = 3y - 2 → 3y = y' + 2 → y = 1/3 y'  + 2/3

y =  x + 2  → 1/3 y' + 2/3 = 1/3 x' + 5/3 || X3
y' + 2  = x'  + 5
y = x + 5 -2
bayangan y = x + 3 memotong sb x , y = 0 → x = - 3
titk (x,y)= (-3, 0)
absis = - 3