Persamaan garis lurus melalui titik A(2,1) dan B(4,7)
Rumus umumnya [tex] \boxed{\boxed{\bf{\frac{y-y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}}}}[/tex]
[tex] \Rightarrow \tt \frac{y-y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} [/tex]
[tex] \Rightarrow \tt \frac{y-1}{7-1} = \frac{x - 2}{4-2} [/tex]
[tex] \Rightarrow \tt \frac{y-1}{6} = \frac{x - 2}{2} [/tex]
[tex] \Rightarrow \tt 2(y - 1) = 6(x - 2)[/tex]
[tex] \Rightarrow \tt 2y - 2 = 6x - 12[/tex]
[tex] \Rightarrow \tt -6x + 2y = -12 + 2[/tex]
[tex] \Rightarrow \tt -6x + 2y = -10[/tex]
[tex] \Rightarrow \tt 6x - 2y = 10[/tex]
[tex] \Rightarrow \tt 3x - y = 5[/tex]
3x - y = 5
3x - y - 5 = 0
y = 3x - 5
.
Maka, persamaan garis lurus yang melewati titik A(2,1) dan B(4,7) adalah 3x - y = 5
@BeruangGemoy :•
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Soal A.
Persamaan garis lurus melalui titik A(2,1) dan B(4,7)
Rumus umumnya [tex] \boxed{\boxed{\bf{\frac{y-y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}}}}[/tex]
Penyelesaian :
[tex] \Rightarrow \tt \frac{y-y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} [/tex]
[tex] \Rightarrow \tt \frac{y-1}{7-1} = \frac{x - 2}{4-2} [/tex]
[tex] \Rightarrow \tt \frac{y-1}{6} = \frac{x - 2}{2} [/tex]
[tex] \Rightarrow \tt 2(y - 1) = 6(x - 2)[/tex]
[tex] \Rightarrow \tt 2y - 2 = 6x - 12[/tex]
[tex] \Rightarrow \tt -6x + 2y = -12 + 2[/tex]
[tex] \Rightarrow \tt -6x + 2y = -10[/tex]
[tex] \Rightarrow \tt 6x - 2y = 10[/tex]
[tex] \Rightarrow \tt 3x - y = 5[/tex]
3x - y = 5
3x - y - 5 = 0
y = 3x - 5
.
Kesimpulan :
Maka, persamaan garis lurus yang melewati titik A(2,1) dan B(4,7) adalah 3x - y = 5
.
@BeruangGemoy :•