Question

1) persamaan garis yang melalui titik (0,6) dan (12,0) adalah?

2) persamaan garis yang melalui titik (2,-5) dan (-3,4) adalah?

Answer 1

1). x1 = 0, y1 = 6
x2 = 12, y2 = 0

y-y1/x-x1 = y2-y1/x2-x1
y-6/x-0 = 0-6/12-0
y-6/x-0 = -6/12
-6(x-0) = 12(y-6)
-6x = 12y-72
-6x-12y = -72
atau
-6x-12y+72 = 0

2). x1 = 2, y1 = -5
x2 = -3, y2 = 4

y-y1/x-x2 = y2-y1/x2-x1
y-(-5)/x-(-3) = 4-(-5)/-3-2
y+5/x+3 = 9/-5
9(x+3) = -5(y+5)
9x+27 = -5y-25
9x+5y = -25-27
9x+5y = -52
atau
9x+5y+52 = 0