Persamaan garis melalui titik (3, 2) dan (-4, 5). Adalah

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

rumus nya = y-y1/y2-y1 = x-x1/x2-x1

y-2/5-2 = x-3/-4-3

y-2/3 = x-3/-7

-7(y-2) = 3(x-3)

-7y+14 = 3x-9
-7y = 3x-9-14
-7y = 3x-23

persamaan garis lurus nya =

3x+7y-23 = 0

semoga membantu

[tex] \frac{y-y1}{y2-y1} = \frac{x-x1}{x2-x1} [/tex]
[tex] \frac{y-2}{5-2} = \frac{x-3}{-4-3} [/tex]
[tex] \frac{y-2}{3} = \frac{x-3}{-7} [/tex]
3x-9 = -7y+14
3x+7y-9+14 = 0
3x+7y+5=0