Question

Fan + Medal

Answer 1

(2, -3)
(5, -3)
(7, -1)
(2, -1)
(-2, -0.5)

Answer 2

hint: only 2 of those options are likely

Answer 3

Can we eliminate E?

Answer 4

I think (7,-1)
explanation:
suppose to have two points belonging to the same line as the drawing below:

we know that
\[\frac{ AP }{ PB }=\frac{ 5 }{ 3 }\]
noe in order to find the x-coordinate of P, we must require that:
measure (AP)/measure(PB)=5/3, so:
\[\frac{ x-x _{A} }{ x _{B}-x }=\frac{ 5 }{ 3 }\]
where x is the x-coordinate of P.
From the above equation, we get:
\[x=\frac{ 3*x _{A} +5*x _{B}}{ 8 }=7\]
analogously for y-coordinate of P, we get:
\[y=\frac{ 3*y _{A} +5*y _{B}}{ 8 }=-1\]

Answer 5

first u have to find the co-ordinates of point D(x,y)

\(\large\tt \begin{align} \color{black}{D(x,y)\\~\\
x=\dfrac{k_1x_b+k_2x_a}{k_1+k_2}\\~\\
=\dfrac{5\times 10+3\times 2}{5+3}\\~\\
=\dfrac{56}{7}\\~\\
=8\\~\\
similarly \\~\\
y=\dfrac{k_1y_b+k_2y_a}{k_1+k_2}\\~\\
=\dfrac{5\times 2+3\times -6}{5+3}\\~\\
=\dfrac{-8}{8}\\~\\
=-1\\~\\
\huge D(x,y)=(8,-1)\\~\\}\end{align}\)
let C(x_1,y_1)
\(\large\tt \begin{align} \color{black}{\sqrt{(x_1-2)^2+(y_1+6)^2}+\sqrt{(x_1-8)^2+(y_1+1)^2}=\sqrt{(8-2)^2+(-1+6)^2}\\~\\
\sqrt{(x_1-2)^2+(y_1+6)^2}+\sqrt{(x_1-8)^2+(y_1+1)^2}=\sqrt{61}\\~\\}\end{align}\)


further u have to use trial and error through options

Answer 6

Thank you all so much!

Answer 7

@CaseyCarns thank you!

Answer 8

*that will be(7,-1)

Answer 9

that is D...they want C

Answer 10

yes i know i just pointed out the correction

Answer 11

I'm just making sure that CaseyCarns knows that the final answer is not (7,-1)

Answer 12

lol

Answer 13

cuz its the co-ordinates of point D right ,not C?

Answer 14

they want point C not point D

Answer 15

@Zarkon Sorry, Sorry, I understand, sorry again!

Answer 16

so its easy now (2,-1)

Answer 17

@CaseyCarns sorry, your solution is (2,-1), not (7,-1), I have made an error, sorry again!

Answer 18

as the y co-ordinate will remain unchanged