1. **State the problem:** We are given two lines $x+y=0$ and $x-y=0$ intersecting at the origin, which is the circumcenter of a right triangle with vertex $A(5,7)$. 2. **Identify the triangle vertices and sides:** The triangle has vertices $A(5,7)$, $B(-7,-5)$, and $C(7,5)$. The sides are $AB$ with slope 1, $AC$ with slope -1, and $BC$ joining points $B$ and $C$. 3. **Find the midpoint of side $BC$:** $$\text{Midpoint } M = \left(\frac{-7+7}{2}, \frac{-5+5}{2}\right) = (0,0)$$ 4. **Find the slope of side $BC$:** $$m_{BC} = \frac{5 - (-5)}{7 - (-7)} = \frac{10}{14} = \frac{5}{7}$$ 5. **Find the slope of the perpendicular bisector of $BC$:** The perpendicular slope is the negative reciprocal: $$m_{\perp} = -\frac{7}{5}$$ 6. **Equation of the perpendicular bisector of $BC$ passing through midpoint $M(0,0)$:** $$y = -\frac{7}{5}x$$ 7. **Perpendicular bisector of $AB$:** - Slope of $AB$ is given as 1. - Midpoint of $AB$: $$\left(\frac{5 + (-7)}{2}, \frac{7 + (-5)}{2}\right) = \left(-1, 1\right)$$ - Perpendicular slope: $$m_{\perp} = -1$$ - Equation: $$y - 1 = -1(x + 1) \Rightarrow y = -x$$ 8. **Perpendicular bisector of $AC$:** - Slope of $AC$ is -1. - Midpoint of $AC$: $$\left(\frac{5 + 7}{2}, \frac{7 + 5}{2}\right) = \left(6, 6\right)$$ - Perpendicular slope: $$m_{\perp} = 1$$ - Equation: $$y - 6 = 1(x - 6) \Rightarrow y = x$$ 9. **Summary:** - Perpendicular bisector of $BC$: $$y = -\frac{7}{5}x$$ - Perpendicular bisector of $AB$: $$y = -x$$ - Perpendicular bisector of $AC$: $$y = x$$ These lines intersect at the circumcenter, which is the origin $(0,0)$, confirming the problem statement. **Final answer:** The perpendicular bisectors are $$y = -\frac{7}{5}x$$, $$y = -x$$, and $$y = x$$ intersecting at the origin.