Question

On a treasure map with coordinates, James is standing on the point (-9, -1). He knows that a treasure chest is located 10 units away from where he is standing and that the coordinates of the treasure chest are such that the x-coordinate is six less than the y-coordinate. Determine and state the possible location(s) of the treasure chest.

Answer 1

To find the possible locations, first consider the following general formula for a circumference of radius r:

$(x-h)^2+(y-k)^2=r^2$

where r is the radius of the circle, and (h,k) is the center of the circle. In this case, the center of the circle is (-9,-1) and the coordinates of the circumference would be the possible locations for the treassure. The radius is 10.

Furthermore, consider that x-coordinate of the treasure is 6 less than y-coordinate, that is:

x = y - 6

Replace the previous expression into the formula for the circumference, with h = -9,

k = -1 and r = 10. Next, order the result to obtain a quadratic equation for y:

$\begin{gathered} (x-(-9))^2+(y-(-1))^2=10^2 \\ (x+9)^2+(y+1)^2=100 \\ (y-6+9)^2+(y+1)^2=100 \\ (y+3)^2+(y-1)^2=100 \\ y^2+6y+9+y^2-2y+1=100 \\ 2y^2+4y+10-100=0 \\ 2y^2+58y-90=0 \\ y^2+29y-45=0 \end{gathered}$

Then, you get a quadratic equation for y. To solve it, use the quadratic formula:

$\begin{gathered} y=\frac{-29\pm\sqrt[]{29^2-4(1)(-45)}}{2} \\ y=\frac{-29\pm\sqrt[]{1021}}{2}=\frac{-29\pm31.95}{2} \\ y_1\approx-30.47 \\ y_2\approx1.47 \end{gathered}$

Then, the solutions for x are:

x1 = y1 - 6 = -30.47 - 6 = -36.47

x2 = y2 - 6 = 1.47 - 6 = -4.53

Hence, the possible locations for the treasure wold be at points (-36.47 , -30.47) and

(-4.53 , 1.47)