Tolong bantu jawab! tentukan persamaan garis yang melalui Pasangan titik berikut: a. A(3,5) dan b(7,11) b. p(4-2) dan q(-6,-12) c. m(-10,-7) dan n(-2,5)

a). a (3,5) dan b (7,11)
x₁ = 3 ; y₁ = 5
x₂ = 7 ; y₂ = 11

[tex] \frac{y-y_1}{y_2-y_1}= \frac{x-x_1}{x_2-x_1} \\ \\ \frac{y-5}{11-5}= \frac{x-3}{7-5} \\ \\ \frac{y-5}{6}= \frac{x-3}{2} \\ \\2(y-5)=6(x-3) \\ 2y-10=6x-18 \\ 6x-2y=18-10 \\ \\ 6x-2y=8 [/tex]

b). p (4, -2) dan q (-6, -12)
x₁ = 4 ; y₁ = -2
x₂ = -6 ; y₂ = -12

[tex]\frac{y-y_1}{y_2-y_1}= \frac{x-x_1}{x_2-x_1} \\ \\ \frac{y-(-2)}{(-5)-(-2)}= \frac{x-4}{(-6)-4} \\ \\ \frac{y+2}{(-3)}= \frac{x-4}{(-10)} \\ \\(-10)(y+2)=(-3)(x-4) \\ -10y-20=-3x+12 \\ 3x-10y=-20+12 \\ \\ 3x-10y=-8[/tex]

c). m (-10, -7) dan n (-2, 5)
x₁ = -10 ; y₁ = -7
x₂ = -2 ; y₂ = 5

[tex]\frac{y-y_1}{y_2-y_1}= \frac{x-x_1}{x_2-x_1} \\ \\ \frac{y-(-7)}{5-(-7)}= \frac{x-(-10}{(-2)-(-10)} \\ \\ \frac{y+7}{(12)}= \frac{x+10}{(8)} \\ \\(8)(y+7)=(12)(x+10) \\ 8y+56=12x+120 \\ 12x-8y=-120+56 \\ \\ 12x-8y=-64[/tex]