tentukan persamaan garis melalui titik (2,3) dan (-2,1)

[tex]\displaystyle \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1} \\ \\ \frac{y-3}{1-3}=\frac{x-2}{-2-2} \\ \\ \frac{y-3}{-2}=\frac{x-2}{-4} \\ \\ -4(y-3)=-2(x-2) \\ 2(y-3)=x-2 \\ 2y-6=x-2 \\ x-2y-2+6=0 \\ x-2y+4=0[/tex]

pakai rumus :
(y-y1) / (y2-y1) = (x-x1) / (x2-x1)

(2,3) (-2,1)
2= x1
3= y1
-2= x2
1=y2

berarti tinggal masukkin rumus

(y-3) / (1-3) = (x-2) / (-2-2)
(y-3) / -2 = (x-2) / -4
-4y+12 = -2x+4
-4y+2x= -8
dikali minus biar y positif
4y - 2x = 8