Titik P(2m,1) terletak pada garis x-6y+ 8 = 0 A.Tentukan m dan koordinat titik p b.Tentukan persamaan garis yang melalui titip P dan sejajar dengan garis 2x-5y-10. #tolong
vhetha
A) x - 6y + 8 = 0 2m - 6(1) + 8 = 0 2m = 6 - 8 2m = -2 m = -1 P(-2,1)
b) 2x - 5y - 10 = 0 -5y = - 2x + 10 y = 2/5 x - 2 m = 2/5 krn sejajar m1 = m2 = 2/5 y = m(x - x1) + y1 y = 2/5(x + 2) - 1 y = 2/5 x + 4/5 - 1 --->×5 5y = 2x - 1
2m - 6(1) + 8 = 0
2m = 6 - 8
2m = -2
m = -1
P(-2,1)
b) 2x - 5y - 10 = 0
-5y = - 2x + 10
y = 2/5 x - 2
m = 2/5
krn sejajar m1 = m2 = 2/5
y = m(x - x1) + y1
y = 2/5(x + 2) - 1
y = 2/5 x + 4/5 - 1 --->×5
5y = 2x - 1