Question:

Step 1: Define line AB and point B. Let $A \equiv (1, 1)$. Let the slope of line AB be $m$. The equation of line AB is given by the point-slope form: $$y - 1 = m(x - 1)$$ Point B is the x-intercept of line AB, so we set $y = 0$: $$0 - 1 = m(x_B - 1)$$ $$-1 = m x_B - m$$ $$m x_B = m - 1$$ $$x_B = 1 - \frac{1}{m}$$ Thus, the coordinates of point B are $\left(1 - \frac{1}{m}, 0\right)$. Step 2: Define line AC and point C. Line AC is perpendicular to line AB. Therefore, the slope of AC is $-\frac{1}{m}$. The equation of line AC, passing through $A(1, 1)$, is: $$y - 1 = -\frac{1}{m}(x - 1)$$ Point C is the y-intercept of line AC, so we set $x = 0$: $$y_C - 1 = -\frac{1}{m}(0 - 1)$$ $$y_C - 1 = \frac{1}{m}$$ $$y_C = 1 + \frac{1}{m}$$ Thus, the coordinates of point C are $\left(0, 1 + \frac{1}{m}\right)$. Step 3: Find the midpoint P of BC. Let $P \equiv (h, k)$ be the midpoint of the line segment BC. Using the midpoint formula: $$h = \frac{x_B + x_C}{2} = \frac{\left(1 - \frac{1}{m}\right) + 0}{2} = \frac{1 - \frac{1}{m}}{2}$$ $$k = \frac{y_B + y_C}{2} = \frac{0 + \left(1 + \frac{1}{m}\right)}{2} = \frac{1 + \frac{1}{m}}{2}$$ Step 4: Determine the locus of P. From the expressions for $h$ and $k$, we have: $$2h = 1 - \frac{1}{m} \quad (*)$$ $$2k = 1 + \frac{1}{m} \quad (**)$$ To eliminate the parameter $m$, we add equations $(*)$ and $(**)$: $$2h + 2k = \left(1 - \frac{1}{m}\right) + \left(1 + \frac{1}{m}\right)$$ $$2h + 2k = 1 - \frac{1}{m} + 1 + \frac{1}{m}$$ $$2h + 2k = 2$$ Dividing by 2, we get: $$h + k = 1$$ Replacing $(h, k)$ with $(x, y)$ to represent the locus of point P, the equation is: $$x + y = 1$$