tolong yaaa pake cara :))))) pake cara

Cara #1

[tex] \frac{c}{d} = \frac{2}{5} \\ \\ 5c = 2d \to c = \frac{2}{5}d \\ \\ x_{1}=c-1= \frac{2}{5}d-1 \\ \\ x_{2}=2c-1= 2.(\frac{2}{5}d)-1 =\frac{4}{5}d-1 \\ \\ y_{1}=d+4 \\ y_{2}=3d+4\\ \\ \\ m= \frac{ y_{2}-y_{1}}{ x_{2}-x_{1}} \\ \\ m= \frac{ (3d+4)-(d+4)}{ (\frac{4}{5}d-1)-(\frac{2}{5}d-1)} \\ \\m= \frac{3d+4-d-4}{\frac{4}{5}d-1-\frac{2}{5}d+1} \\ \\m= \frac{2d}{\frac{2}{5}d} \\ \\ m = (2).(\frac{5}{2}) \\ \\ m=5[/tex]

Cara #2

[tex]\frac{c}{d} = \frac{2}{5}.....maka\ c =2\ dan\ d=5 \\ \\ x_{1}=c-1= 2-1 = 1 \\y_{1}=d+4=5+4=9 \\ x_{2}=2c-1= 2(2)-1=4-1=3\\ y_{2}=3d+4=3(5)+4=15+4=19 \\ \\ m= \frac{ y_{2}-y_{1}}{ x_{2}-x_{1}} \\ \\ m= \frac{ 19-9}{ 3-1} \\ \\ m= \frac{10}{2} \\ \\ m=5[/tex]

Jadi gradien garis k adalah 5..........Jawaban (B)