+0

0
55
5
+355

I hope this link works now

waffles  Oct 16, 2017
edited by waffles  Oct 16, 2017
edited by waffles  Oct 16, 2017
edited by waffles  Oct 16, 2017

#5
+5241
+1

Let's call the coordinates of the treasure  (x, y) .

The treasure is  10  yards away from  (0, 0) ,

and we know that the distance between any two points  = $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

Using the points  (x, y)  and  (0, 0)  and a distance of  10 , we have...

10  =  $$\sqrt{(x-0)^2+(y-0)^2}\,=\,\sqrt{x^2+y^2}$$                         →     100  =  x2 + y2

The treasure is  10  yards from  (15, 0) . So we know that

10  =  $$\sqrt{(x-15)^2+(y-0)^2}\,=\,\sqrt{(x-15)^2+y^2}$$          →     100  =  (x - 15)2 + y2

Now we have two equations and two variables.

They both equal 100, so we can set these equal.

x2 + y2  =  (x - 15)2 + y2

x2  =  (x - 15)2

x  =  ±(x - 15)

x  =  x - 15          or          x  =  -(x - 15)

0  =  -15                           x  =  -x + 15

false                              2x  =  15

x  =  15 / 2

Plug this value for  x  into  x2 + y2 = 100  to find the value of  y .

(15 / 2)2 + y2  =  100

y2  =  100 -  (15 / 2)2

y   =  ±√[ 100 -  (15 / 2)2 ]

y  ≈  6.6

Since there are only positive values of  y  on the map, the coordinates are  ≈  (7.5, 6.6)

hectictar  Oct 16, 2017
Sort:

#1
+91005
0

That link does not work for me waffles. :(

I did answer one of  your questions down the page a bit.

I 'guessed' what the pic might be :)    Did I get the pic right ??

Melody  Oct 16, 2017
#2
+355
0

I posted a new photo

waffles  Oct 16, 2017
#3
+91005
0

It would  be nice if you at least commented,

Melody  Oct 16, 2017
#4
+91005
0

Thanks forthe comment Waffles :))

Melody  Oct 16, 2017
#5
+5241
+1

Let's call the coordinates of the treasure  (x, y) .

The treasure is  10  yards away from  (0, 0) ,

and we know that the distance between any two points  = $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

Using the points  (x, y)  and  (0, 0)  and a distance of  10 , we have...

10  =  $$\sqrt{(x-0)^2+(y-0)^2}\,=\,\sqrt{x^2+y^2}$$                         →     100  =  x2 + y2

The treasure is  10  yards from  (15, 0) . So we know that

10  =  $$\sqrt{(x-15)^2+(y-0)^2}\,=\,\sqrt{(x-15)^2+y^2}$$          →     100  =  (x - 15)2 + y2

Now we have two equations and two variables.

They both equal 100, so we can set these equal.

x2 + y2  =  (x - 15)2 + y2

x2  =  (x - 15)2

x  =  ±(x - 15)

x  =  x - 15          or          x  =  -(x - 15)

0  =  -15                           x  =  -x + 15

false                              2x  =  15

x  =  15 / 2

Plug this value for  x  into  x2 + y2 = 100  to find the value of  y .

(15 / 2)2 + y2  =  100

y2  =  100 -  (15 / 2)2

y   =  ±√[ 100 -  (15 / 2)2 ]

y  ≈  6.6

Since there are only positive values of  y  on the map, the coordinates are  ≈  (7.5, 6.6)

hectictar  Oct 16, 2017

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