Find the coordinates of a point equidistant from $P_1(1,-6)$, $P_2(5,-6)$, and $P_3(6,-1)$. (2,-2) (3,-2) (3,-3) (2,-3) Click to Show/Hide Solution \[ d_1 = d_2 = d_3 \] \[ d_1 = d_2 \] \[ \sqrt{(1-x)^2 + (-6-y)^2} = \sqrt{(5-x)^2 + (-6-y)^2} \] \[ (1-x)^2 + (-6-y)^2 = (5-x)^2 + (-6-y)^2 \] \[ (1-x)^2 = (5-x)^2 \] \[ (1-x)^2 = (5-x)^2 \] \[ 1 - 2x + x^2 = 25 - 10x + x^2 \] \[ 8x = 24 \] \[ x = 3 \] \[ d_2 = d_3 \] \[ \sqrt{(5-x)^2 + (-6-y)^2} = \sqrt{(6-x)^2 + (-1-y)^2} \] \[ (5-x)^2 + (-6-y)^2 = (6-x)^2 + (-1-y)^2 \] \[ 25 - 10x + x^2 + 36 + 12y + y^2 = 36 - 12x + x^2 + 1 + 2y + y^2 \] \[ 2x + 10y = -24 \] \[ 2(3) + 10y = -24 \] \[ 10y = -30 \] \[ y = -3 \] $ \therefore P(3,-3) \leftarrow \text{answer} $
\[
d_1 = d_2 = d_3
\]
\[
d_1 = d_2
\]
\[
\sqrt{(1-x)^2 + (-6-y)^2} = \sqrt{(5-x)^2 + (-6-y)^2}
\]
\[
(1-x)^2 + (-6-y)^2 = (5-x)^2 + (-6-y)^2
\]
\[
(1-x)^2 = (5-x)^2
\]
\[
(1-x)^2 = (5-x)^2
\]
\[
1 – 2x + x^2 = 25 – 10x + x^2
\]
\[
8x = 24
\]
\[
x = 3
\]
\[
d_2 = d_3
\]
\[
\sqrt{(5-x)^2 + (-6-y)^2} = \sqrt{(6-x)^2 + (-1-y)^2}
\]
\[
(5-x)^2 + (-6-y)^2 = (6-x)^2 + (-1-y)^2
\]
\[
25 – 10x + x^2 + 36 + 12y + y^2 = 36 – 12x + x^2 + 1 + 2y + y^2
\]
\[
2x + 10y = -24
\]
\[
2(3) + 10y = -24
\]
\[
10y = -30
\]
\[
y = -3
\]
$ \therefore P(3,-3) \leftarrow \text{answer} $
