Two friends work at different buildings on opposite sides of the city. They would like to meet for coffee after work someplace in between both locations. One friend works 9 blocks west and 6 blocks south of the center of town. The other friend works 9 blocks east and 4 blocks north of the center of town. Which restaurant is closest to the midpoint of the two locations?
A. Restaurant Casa, which is 1 block south of the center of town B. Pizza di Franco, which is 9 blocks north and 5 blocks west of the center of town C. Cafe Rosa, which is 2 blocks north and 2 blocks east of the center of town D. 7th St. Market, which is 2 blocks south and 3 blocks east of the center of town
Do you have a picture?
no
can yu still help tho?
I am trying to solve the problem.
Do you know what coordinate is the center town?
no sir
You don't have to call me sir. I am not one.
oh okay lol
Are you learning slopes?
distance and midpoint
I am just asking just in case if there is anything that this problem is related to.
not slopes
One minute, I am almost getting there.
Do you know the Pythagorean Theorem?
yes a^2+b^2=c^2
Great. I have put an attachment. You can see that the coordinate grids can show it as a triangle. Find the distance of both the legs to get a and b in the pythagorean theorem.
Got it?
no...
You can see a triangle right?
yes
What I have just did was connect the two coordinate dots.
can i have the coordinates
or what ever i need to put in the formua
Well I got it as (9, -6) and (9, 4). Let's consider the center town that have the coordinates of (0, 0). If you move nine spaces both to the left and to the right or six spaces down or four spaces up.
To find d, which is basically the hypotenuse, you need to figure out a and b. That is the distance of both of the legs.
wait are yu saying i need to use the distance formula
You need to use the Pythagorean Theorem.
but thats a+b
Yes, you just need to find the number of units.
That way you can replace a and b for a value.
this sounds like d= (x1-x2) + (y1-y2)
Yes, or \[\\ d=\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}} \\\]
yeah but
The easiest way is the Pythagorean Theorem in my opinion.
Find the units in A and B.
Leg A has 18 units Leb B has 10 units. \[18^2 + 10^2 = c^2\]
oh okey
is it 36+20=c^2
so 56=c^2
18 x 18 = 324 10 x 10 = 100 324 + 100 = 424 Square root of 424 ~ 20.59 ~ 21
holy cow
but where are the corrdinates
this didnt solve anything
You don't need them in the Pythagorean Theorem.
how do i get my answer
So let's say 20 is the midpoint.
40 and -40
If you look closely then it shows that the -1, under the 0, is the closest to the midpoint.
Looking at this from a general point of view.
I would say that this would be A.
okay? but yu cant walk negative blocks
Not really.
But I never mentioned -40 though.
Hopefully A is the answer. I just learned this new with you.
yeah but yu said midpoint idk
Yeah
oh okay well thank yu
Midpoint of -40 and 40 = 0, not 20. Midpoint of 0 and 40 = 20
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