Persamaan garis yang melalui titik ( 6,-2 ) dan tegak garis dengan persamaan 3x - 2y + 6 = 0 adalah... a. 3y -2x + 18 = 0 b. 3y +2x - 6 = 0 c. 2y - 3x +22 =0 d. 2y + 3x +8 =0
ekamiaww
Ax + by + c = 0 --> m = -a/b 3x - 2y + 6 = 0 --> m₁ = -3/-2 = 3/2 tegak lurus maka, m₂ = -2/3 persamaan garis dgn m = -2/3 dan melalui titik (6,-2): y - y₁ = m (x - x₁) y - (-2) = -2/3 (x - 6) y + 2 = -2/3 (x - 6) --> kedua ruas dikali 3 3y + 6 = -2 (x - 6) 3y + 6 = -2x + 12 2x + 3y + 6 - 12 = 0 2x + 3y - 6 = 0 atau 3y + 2x - 6 = 0 (B)
3x - 2y + 6 = 0 --> m₁ = -3/-2 = 3/2
tegak lurus maka, m₂ = -2/3
persamaan garis dgn m = -2/3 dan melalui titik (6,-2):
y - y₁ = m (x - x₁)
y - (-2) = -2/3 (x - 6)
y + 2 = -2/3 (x - 6) --> kedua ruas dikali 3
3y + 6 = -2 (x - 6)
3y + 6 = -2x + 12
2x + 3y + 6 - 12 = 0
2x + 3y - 6 = 0
atau
3y + 2x - 6 = 0 (B)
semoga membantu ya :)
3x - 2y = -6
ax + by = c
a = 3
b = -2
c = -6
m₁ = -a/b
= -3/-2
= 3/2
m₁ × m₂ = -1
3/2 × m₂ = -1
m₂ = -2/3
y - y₁ = m(x - x₁)
y - (-2) = -2/3(x - 6)
y + 2 = -2/3x + 4
3y + 6 = -2x + 12
3y + 2x + 6 - 12 = 0
3y + 2x - 6 = 0
Jadi jawabannya adalah yang B. 3y + 2x - 6 = 0