persamaan garis yg melalui titik (4,-3) dan tegak lurus dgn garis 4y-6x+10=0

Kelas 8 Matematika
Bab Persamaan Garis

ax + by + c = 0, tegak lurus, (p, q)
bx - ay - (bp - aq) = 0

-6x + 4y + 10 = 0, tegak lurus, (4, -3)
4x + 6y - (4 . 4 + 6 . (-3)) = 0
4x + 6y - (16 - 18) = 0
4x + 6y - (-2) = 0
2x + 3y + 1 = 0

Garis tegak lurus :
m1 × m2 = -1

4y - 6x + 10 = 0
Rumus mencari gradiennya :
-a/b
= -(-6)/4
= 6/4
= 3/2
Catatan :
a = angka di depan variabel x
b = angka di depan variabel y

m1 x m2 = -1
3/2 x m2 = -1
3 m2 = -1 (2)
3 m2 = -2
m2 = -2/3

( 4,-3 )
x1 , y1

Rumus mencari persamaan garis melalui satu titik :
y - y1 = m ( x - x1 )
y - (-3) = -2/3 ( x - 4 )
y + 3 = -2/3x + 8/3
--------------------------- ×3
3y + 9 = -2x + 8
3y + 2x = 8 -9
3y + 2x = -1
Persamaan garis :
2x + 3y = -1
atau
2x + 3y + 1 = 0