Key Concepts and Formulas

• Slope of a line: The slope of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
• Perpendicular lines: If two lines are perpendicular, the product of their slopes is -1, i.e., $m_1 \cdot m_2 = -1$.
• Point-slope form of a line: The equation of a line passing through point $(x_1, y_1)$ with slope $m$ is given by $y - y_1 = m(x - x_1)$.
• Orthocenter: The point of intersection of the altitudes of a triangle.

Step-by-Step Solution

Step 1: Find the equation of the altitude from vertex C to side AB.

• 1a. Calculate the slope of side AB. We calculate the slope of side AB because the altitude from C will be perpendicular to this side. Using $A(-1, 7)$ and $B(-7, 1)$: $m_{AB} = \frac{1 - 7}{-7 - (-1)} = \frac{-6}{-6} = 1$

• 1b. Calculate the slope of the altitude from C to AB. Since the altitude from C is perpendicular to AB, we find the negative reciprocal of $m_{AB}$. Let the altitude from C be $CD$. $m_{CD} = -\frac{1}{m_{AB}} = -\frac{1}{1} = -1$

• 1c. Find the equation of the altitude CD. We use the point-slope form to find the equation of the line passing through $C(5, -5)$ with slope $m_{CD} = -1$: $y - (-5) = -1(x - 5)$ $y + 5 = -x + 5$ $x + y = 0 \quad \text{(Equation 1)}$

Step 2: Find the equation of the altitude from vertex A to side BC.

• 2a. Calculate the slope of side BC. We calculate the slope of side BC because the altitude from A will be perpendicular to this side. Using $B(-7, 1)$ and $C(5, -5)$: $m_{BC} = \frac{-5 - 1}{5 - (-7)} = \frac{-6}{12} = -\frac{1}{2}$

• 2b. Calculate the slope of the altitude from A to BC. Since the altitude from A is perpendicular to BC, we find the negative reciprocal of $m_{BC}$. Let the altitude from A be $AE$. $m_{AE} = -\frac{1}{m_{BC}} = -\frac{1}{-1/2} = 2$

• 2c. Find the equation of the altitude AE. We use the point-slope form to find the equation of the line passing through $A(-1, 7)$ with slope $m_{AE} = 2$: $y - 7 = 2(x - (-1))$ $y - 7 = 2(x + 1)$ $y - 7 = 2x + 2$ $2x - y = -9 \quad \text{(Equation 2)}$

Step 3: Find the intersection point of the two altitudes.

We solve the system of equations formed by Equation 1 and Equation 2 to find the coordinates of the orthocenter. Equation 1: $x + y = 0$ Equation 2: $2x - y = -9$

Adding the two equations: $(x + y) + (2x - y) = 0 + (-9)$ $3x = -9$ $x = -3$

Substituting $x = -3$ into Equation 1: $-3 + y = 0$ $y = 3$

Step 4: State the coordinates of the orthocenter.

The intersection point, which is the orthocenter, is $(-3, 3)$.

Common Mistakes & Tips:

• Sign Errors: Pay close attention to signs when calculating slopes and using the point-slope form.
• Perpendicular Slope: Remember that the slope of a perpendicular line is the negative reciprocal.
• Using the Correct Vertex: When finding the equation of an altitude, make sure to use the coordinates of the vertex from which the altitude is drawn.

Summary

To find the orthocenter, we calculated the equations of two altitudes of the triangle and then found their point of intersection. This was done by finding the slopes of two sides, finding the slopes of the perpendicular altitudes, and then using the point-slope form to determine the equations of the altitudes. Solving the system of equations gave us the coordinates of the orthocenter. The orthocenter of $\Delta ABC$ with vertices $A(-1, 7)$, $B(-7, 1)$, and $C(5, -5)$ is $(-3, 3)$.

Final Answer The final answer is \boxed{(-3, 3)}, which corresponds to option (A).