Step 1: Understanding the Orthocenter

The orthocenter (P) of a triangle is the intersection of its three altitudes. An altitude is a line passing through a vertex and perpendicular to the opposite side.

We will:

1. Find the slopes of sides AB, BC, and AC.
2. Determine the perpendicular slopes to find the equations of the altitudes.
3. Solve for their intersection (orthocenter).

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Step 2: Find the Slopes of Sides

Using the slope formula:
$m = \frac{y_2 - y_1}{x_2 - x_1}$

Slope of AB:

$m_{AB} = \frac{6 - (-2)}{0 - (-6)} = \frac{6 + 2}{0 + 6} = \frac{8}{6} = \frac{4}{3}$

Slope of BC:

$m_{BC} = \frac{0 - 6}{6 - 0} = \frac{-6}{6} = -1$

Slope of AC:

$m_{AC} = \frac{0 - (-2)}{6 - (-6)} = \frac{0 + 2}{6 + 6} = \frac{2}{12} = \frac{1}{6}$

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Step 3: Find the Equations of Altitudes

The perpendicular slope to a line $y = mx + b$ is $m_{\perp} = -\frac{1}{m}$.

Equation of Altitude AP (Perpendicular to BC)

• $m_{BC} = -1$
• Perpendicular slope $m_{\perp} = 1$
• Passes through $A(-6, -2)$
• Using point-slope form: $y - y_1 = m(x - x_1)$ $y - (-2) = 1(x - (-6))$ $y + 2 = x + 6$ $y = x + 4$

Equation of Altitude BP (Perpendicular to AC)

• $m_{AC} = \frac{1}{6}$
• Perpendicular slope $m_{\perp} = -6$
• Passes through $B(0,6)$
• Using point-slope form: $y - 6 = -6(x - 0)$ $y - 6 = -6x$ $y = -6x + 6$

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Step 4: Find the Intersection of AP and BP (Orthocenter P)

Solve the system: $y = x + 4$ $y = -6x + 6$

Set equations equal to each other: $x + 4 = -6x + 6$

Solve for $x$: $x + 6x = 6 - 4$ $7x = 2$ $x = \frac{2}{7}$

Substituting $x = \frac{2}{7}$ into $y = x + 4$: $y = \frac{2}{7} + 4 = \frac{2}{7} + \frac{28}{7} = \frac{30}{7}$

Thus, the orthocenter $P$ is: $\left( \frac{2}{7}, \frac{30}{7} \right)$

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Final Answers

• Equation of Altitude AP: $y = x + 4$
• Equation of Altitude BP: $y = -6x + 6$
• Orthocenter P: $\left( \frac{2}{7}, \frac{30}{7} \right)$

Would you like a more detailed explanation or a diagram?

Related Questions

1. How do we find the equation of the third altitude (CP)?
2. What is the centroid of the triangle, and how does it relate to the orthocenter?
3. How can we verify that the intersection of two altitudes is sufficient to find the orthocenter?
4. What happens if a triangle is right-angled—where is the orthocenter located?
5. Can the orthocenter ever be outside the triangle? If so, when?

Math Tip:

For any right triangle, the orthocenter is simply the vertex at the right angle!