Persamaan garis lurus yg melalui titik p(3,-4)dan Q(-2,6)adalah

diketahui : x1 = 3
x2= -2
y1 = -4
y2 = 6
jawab : y-y1/y2-y1 = x-x1/x2-x1
y-(-4)/6-(-4)= x -3/-2-3
y+4/6+4 = x-3/-2-3
y+4/10 = x-3/-5
-5( y+4 ) = 10( x-3 )
-5y-20 = 10x -30
-5y = 10x-30+20
-5y = 10x-10
10x+5y = 10 atau y = -2x+5