Tentukan persamaan garis yang melalui titik= a. (2,5) dan (4,2) b. (-3,2) dan (4,-1) c. (-2,-3) dan (-5,-4)

a)
[tex] \frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1} \\ \frac{y - 5}{x - 2} = \frac{2 - 5}{4 - 2} \\ \frac{y - 5}{x - 2} = \frac{-3}{2} \\ 2(y - 5) = -3(x - 2) \\ 2y - 10 = -3x + 6 \\ 3x + 2y = 16 [/tex]

b)
[tex] \frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1} \\ \frac{y - 2}{x - (-3)} = \frac{-1 - 2}{4 - (-3)} \\ \frac{y - 2}{x + 3} = \frac{-3}{7} \\ 7(y - 2) = -3(x + 3) \\ 7y - 14 = -3x - 9 \\ 3x + 7y = 5 [/tex]

c)
[tex] \frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1} \\ \frac{y - (-3)}{x - (-2)} = \frac{-4 - (-3)}{-5 - (-2)} \\ \frac{y + 3}{x + 2} = \frac{-1}{-3} \\ -3(y + 3) = -1(x + 2) \\ -3y - 9 = -x - 2 \\ x - 3y = 7 [/tex]

Mohon koreksinya kalau salah

langsung saja....


a)
(y-5)/(2-5) = (x-2)/(4-2)
(y-5)/-3 = (x-2)/2
2y-10 = -3x+6
2y = -3x+6+10
2y = -3x + 16

b)
(y-2)/(-1-2) = (x-(-3))/(4-(-3))
(y-2)/-3 = (x+3)/7
7y-14 = -3x-9
7y = -3x-9+14
7y = -3x + 5

c)
(y-(-3))/(-4-(-3)) = (x-(-2))/(-5-(-2))
(y+3)/-1 = (x+2)/-3
-3y-9 = -x-2
-3y = -x-2+9
-3y = -x+7
3y = x - 7