1.Persamaan garis yang melalui titik (-2,1) dan sejajar dengan garis yang melalui titik (4,3) dan (-2,-5),jelaskan caranya adalah.... A. 3y-4x=5 B. 3y+4y=-11 C.

1.gradien garis= y2-y1 / x2-x1
= -5-3 /-2-4 = -8/-6 = 4/3
sejajar nilai m1 =m2
melalui titik (-2,1)
y-y1= m(x-x1 )
y-1 = 4/3 (x -(-2))
y-1 = 4/3 x +8/3
y= 4/3x + 8/3 +1 ( masing2 dikali 3)
3y= 4x +8 +3
3y -4x = 11

2. gradien garis dari (3,-4) (-1,2)=
= y2-y1 / x2-x1
= 2 - (-4) / -1 - 3
= 6/4 = 3/2
tegak lurus m1 x m2 = -1
3/2 x m2 = -1
m2 = -2/3
melalui ttik(-4,-3) maka persamaannya=
y-y1= m(x-x1 )
= y-(-3) = -2/3 (x- (-4))
= y+3 = -2/3x+ 8/3
y+ 2/3x = 8/3 -3 (di x 3)
3y + 2x = 8 -9
3y +2x = -1

3. pers (0,-4) (5,0)
y-y1 per y2-y1 = x-x1 per x2-x1
y-(-4) / 0-(-4) = x -0 / 5-0
y+4 /4 = x /5
4(x) = 5(y+4)
4x= 5y +20
4x -5y -20 =0

4. 6y-8x=7 gradiennya (m) = -a/b = -6/-8= 3/4
tegak lurus m1 x m2 = -1
3/4 x m2 = -1
m2 = -4/3