How to solve this coordinate problem??

[RandomGuyThatNoobOnMath]

I had no idea how to do this ;(

PLEASE HELP

 

[Otis]

I had no idea how to do this

Hello. The Distance Formula will work, to find the x-coordinate of point K.

They've already given you the y-coordinate of point K. Substitute the values (-2,6) and (x,4) and √104 into the Distance Formula, and then solve the resulting equation for x. NOTE: You'll get two results for x; use the given diagram to pick the right one.

Once you have that, the Midpoint Formula will provide the coordinates of the point halfway between points K and L.

If you need help with either of these steps, please post how far you got. Cheers
[imath]\;[/imath]

 

[RandomGuyThatNoobOnMath]

Hello. The Distance Formula will work, to find the x-coordinate of point K.

They've already given you the y-coordinate of point K. Substitute the values (-2,6) and (x,4) and √104 into the Distance Formula, and then solve the resulting equation for x. NOTE: You'll get two results for x; use the given diagram to pick the right one.

Once you have that, the Midpoint Formula will provide the coordinates of the point halfway between points K and L.

If you need help with either of these steps, please post how far you got. Cheers
[imath]\;[/imath]

I actually also know I can use the distance formula to find it, the answer is wrong every time ?, but I have no idea where I have do it wrong. I always get that X is 10. The picture show above is my formula. ? Can you tell me where do I do wrong ? TQ?

 

[Otis]

Hi. The first line of your work is good, but in the second line we cannot simplify only part of the square root on the left-hand side. In other words, this type of separation is not allowed:

\(\displaystyle \color{red}\sqrt{(x \;– \;\text{-}2)^2} + \sqrt{4}\color{black}\)

Maybe you were thinking of the property
[imath]\quad\sqrt{a\times b}=\sqrt{a} \times \sqrt{b}[/imath]
but there is no similar property for addition.

Instead, we can remove the square root symbol on each side of line 1) by squaring both sides of the equation. Your line 2) could look like this:

\(\displaystyle {\bigg( \sqrt{(x \;– \;\text{-}2)^2 + (4 \;–\; 6)^2} \bigg)}^2 = {\bigg(\sqrt{104} \bigg)}^2\)

Also, subtracting a negative number is the same thing as adding the opposite number, so line 3) could look like this:

\(\displaystyle (x + 2)^2 + (\text{-}2)^2= 104\)

Continue solving for x. Feel free to ask questions, if anything's unclear.
[imath]\;[/imath]

 

[Steven G]

If sqrt(a^2 + b^2) = sqrt(a^2) + sqrt(b^2) = a + b, then the pythagorean theorem would reduce to a+b=c!

 

[Otis]

c!

You did that on purpose! (Was it because I'd joked that you couldn't find a date?) ?
[imath]\;[/imath]

 

[RandomGuyThatNoobOnMath]

Hi. The first line of your work is good, but in the second line we cannot simplify only part of the square root on the left-hand side. In other words, this type of separation is not allowed:

\(\displaystyle \color{red}\sqrt{(x \;– \;\text{-}2)^2} + \sqrt{4}\color{black}\)

Maybe you were thinking of the property
[imath]\quad\sqrt{a\times b}=\sqrt{a} \times \sqrt{b}[/imath]
but there is no similar property for addition.

Instead, we can remove the square root symbol on each side of line 1) by squaring both sides of the equation. Your line 2) could look like this:

\(\displaystyle {\bigg( \sqrt{(x \;– \;\text{-}2)^2 + (4 \;–\; 6)^2} \bigg)}^2 = {\bigg(\sqrt{104} \bigg)}^2\)

Also, subtracting a negative number is the same thing as adding the opposite number, so line 3) could look like this:

\(\displaystyle (x + 2)^2 + (\text{-}2)^2= 104\)

Continue solving for x. Feel free to ask questions, if anything's unclear.
[imath]\;[/imath]

I did it , the answer is C , TQ So Much ! ?