Question

suppose that receiving stations x, y, and z are located on a coordinate plane at the points (14,3), (-4, -10), and (-2,1), respectively. the epicenter of an earthquake is determined to be 13 units from x, 10 units from y, and 5 units from z. where on the coordinate plane is the epicenter located?
find the coordinates of the epicenter.
□ (type an ordered pair.)

Explanation:

Step1: Use the distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Let the epicenter be \((x,y)\).
The distance from \((x,y)\) to \((14,3)\) is \(\sqrt{(x - 14)^2+(y - 3)^2}=13\), so \((x - 14)^2+(y - 3)^2 = 169\).
The distance from \((x,y)\) to \((-4,-10)\) is \(\sqrt{(x + 4)^2+(y + 10)^2}=10\), so \((x + 4)^2+(y + 10)^2=100\).
The distance from \((x,y)\) to \((-2,1)\) is \(\sqrt{(x + 2)^2+(y - 1)^2}=5\), so \((x + 2)^2+(y - 1)^2 = 25\).

Step2: Expand the equations

Expand \((x - 14)^2+(y - 3)^2 = 169\): \(x^{2}-28x + 196+y^{2}-6y+9 = 169\), or \(x^{2}+y^{2}-28x-6y+36 = 0\).
Expand \((x + 4)^2+(y + 10)^2=100\): \(x^{2}+8x + 16+y^{2}+20y + 100 = 100\), or \(x^{2}+y^{2}+8x+20y + 16 = 0\).
Expand \((x + 2)^2+(y - 1)^2 = 25\): \(x^{2}+4x+4+y^{2}-2y + 1 = 25\), or \(x^{2}+y^{2}+4x-2y-20 = 0\).

Step3: Subtract equations

Subtract the second - expanded equation from the first:
\((x^{2}+y^{2}-28x-6y+36)-(x^{2}+y^{2}+8x+20y + 16)=0\)
\(-36x-26y+20 = 0\), simplify to \(18x + 13y-10 = 0\), so \(y=\frac{10 - 18x}{13}\).
Subtract the third - expanded equation from the second:
\((x^{2}+y^{2}+8x+20y + 16)-(x^{2}+y^{2}+4x-2y-20)=0\)
\(4x+22y+36 = 0\), simplify to \(2x + 11y+18 = 0\).

Step4: Substitute \(y\) into the new equation

Substitute \(y=\frac{10 - 18x}{13}\) into \(2x + 11y+18 = 0\):
\(2x+11\times\frac{10 - 18x}{13}+18 = 0\)
Multiply through by 13: \(26x+110-198x + 234 = 0\)
\(-172x=-344\), so \(x = 2\).

Step5: Find \(y\)

Substitute \(x = 2\) into \(y=\frac{10 - 18x}{13}\), \(y=\frac{10-18\times2}{13}=\frac{10 - 36}{13}=-2\).

Answer:

\((2,-2)\)