Question

Tentukanlah persamaan gar is yang melalui (9,-6) Dan tegak lurus gar is 2x+3y-5=0

Answer 1

M=-2/3 
tegak lurus m1 * m2 = -1 
y-y1 = m(x-x1)
y+6 = 3/2(x-9) 
2y+12 = 3x-27
3x - 2y - 39 = 0

Answer 2

Pers garis melalui (9,-6)
y - y1 = m(x - x1)
y - (-6) = m(x - 9)
y + 6 = mx - 9m ...1)

tegak lurus dgn garis
2x + 3y -5 =0
3y = -2x +5
y =-2/3x +5/3 => bergradien -2/3

karena tegak lurus
m1 x m2 = -1
m1 x -2/3 = -1
m1 = -1/(-2/3)
m1 = -1.3/-2
m1 = 3/2 (gradien garis (9,-6))

substusi dgn pers 1
y + 6 = mx - 9m
y+ 6 = 3/2x - 9.3/2
2y = 3x - 27 -12
2y - 3x + 39 = 0