Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
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.
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
