Question

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Answer 1

\[\huge d = \sqrt{(x _{2}-x _{1})^{2}+(y _{2}-y _{1})^{2}}\]

Answer 2

Use the distance formula

Answer 3

Your c is (x1,y1) and your d is (x2,y2) mind plugging in the numbers in yourself here? So I know you understand it?

Answer 4

Good job :) do that and squareroot it

Answer 5

Wait hold on you did it wrong, you're not suppose to square the inside.

\[\huge d = \sqrt{(13-1)^2+(-2-7)^2}\]

Answer 6

You should get 15.

Answer 7

Nope, that's the final answer, I included the square root for it, try it yourself, and see if you get the same answer as me.

Answer 8

To get midpoint this is the formula

\[\huge \left( \frac{ x _{1}+x _{2} }{ 2 },\frac{ y _{1}+y _{2} }{ 2 }\right)\]\]

Answer 9

To get slope use this formula, yup there are a lot of formulas lol.

\[\huge m = \frac{ y_{2}-y _{1} }{ x _{2}-x _{1} }\] (where m represents the slope).

Answer 10

That doesn't make sense, midpoint gives you two coordinate points, basically you'll get 2 new points (x,y). Hence, the comma in the formula :p

Answer 11

I got (7, 5/2)

Answer 12

You can check if this is correct

Answer 13

(1, 7) and (13, -2) Putting these for reference, so let me try these again.

Answer 14

\[\left( \frac{ 1+13 }{ 2 }, \frac{ -2+7 }{ 2 }\right) => (7, \frac{ 5 }{ 2 })\]

Answer 15

I'm still getting the same answers, maybe you mixed it up, it's alright :)

Answer 16

Alright now lets find the slope

Answer 17

\[m=\frac{ -2-7 }{ 13-1 } =>\frac{ -9 }{ 12 } => \frac{ -3 }{ 4 }\]

Answer 18

Well (h, k) is the center.

Answer 19

\[\huge \frac{ (x-h)^2 }{ a^2 }+\frac{ (y-k)^2 }{ b^2 } = 1\]

This is the formula for a wider than tall ellipse, as your question is asking, here (h,k) represents the center! :)

Answer 20

I think, I've gave you more than enough information to figure this stuff out now :p

Answer 21

The center is (-2,-7)