Answer:
Options C. and D. are correct.
Step-by-step explanation:
Let [tex](x_1,y_1)=(0,5),\,(x_2,y_2)=(2,8)[/tex]
Equation of a line is given by [tex]y-y_1=(\frac{y_2-y_1}{x_2-x_1})(x-x_1)[/tex]
[tex]y-5=(\frac{8-5}{2-0})(x-0)\\\\y-5=\frac{3}{2}x\\\\2y-10=3x\\3x-2y+10=0[/tex]
Put [tex](x,y)=(5,11)[/tex]
[tex]3(5)-2(11)+10=15-22+10=3\neq 0[/tex]
Put [tex](x,y)=(5,10)[/tex]
[tex]3(5)-2(10)+10=15-20+10=5\neq 0[/tex]
Put [tex](x,y)=(6,14)[/tex]
[tex]3(6)-2(14)+10=18-28+10=0[/tex]
Put [tex](x,y)=(30,50)[/tex]
[tex]3(30)-2(50)+10=90-100+10=0[/tex]
Put [tex](x,y)=(40,60)[/tex]
[tex]3(40)-2(60)+10=0\\120-120+10=10\neq 0[/tex]
So, points [tex](6,14),\,(30,50)[/tex] satisfy the equation [tex]3x-2y+10=0[/tex]
Therefore,
points [tex](6,14),\,(30,50)[/tex] lie on the line through [tex](0,5)[/tex] and [tex](2,8)[/tex]
Options C. and D. are correct.
