Key Concepts and Formulas

• Orthocenter: The point of intersection of the altitudes of a triangle.
• Altitude: A line segment from a vertex of a triangle perpendicular to the opposite side.
• Perpendicular Lines: Two lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $m_1 \cdot m_2 = -1$.
• Slope Formula: The slope of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Step-by-Step Solution

Let the vertices of the triangle be $A(0, 2)$, $B(4, 3)$, and $C(a, b)$. The orthocenter is $H(0, 0)$.

Step 1: Use the fact that $CH$ is perpendicular to $AB$

Since $H$ is the orthocenter, the line segment $CH$ is an altitude and is therefore perpendicular to the side $AB$. We will find the slopes of $CH$ and $AB$, and then use the perpendicularity condition.

• Calculate the slope of $AB$ ($m_{AB}$): $m_{AB} = \frac{3 - 2}{4 - 0} = \frac{1}{4}$

• Calculate the slope of $CH$ ($m_{CH}$): $m_{CH} = \frac{b - 0}{a - 0} = \frac{b}{a}$

• Apply the perpendicularity condition ($m_{CH} \cdot m_{AB} = -1$): Since $CH \perp AB$, we have: $m_{CH} \cdot m_{AB} = -1$ $\frac{b}{a} \cdot \frac{1}{4} = -1$ $\frac{b}{4a} = -1$ $b = -4a$ Thus, we have our first equation: $4a + b = 0 \quad \ldots(1)$

Step 2: Use the fact that $BH$ is perpendicular to $AC$

The line segment $BH$ is an altitude and is therefore perpendicular to the side $AC$. We will find the slopes of $BH$ and $AC$, and then use the perpendicularity condition.

• Calculate the slope of $AC$ ($m_{AC}$): $m_{AC} = \frac{b - 2}{a - 0} = \frac{b - 2}{a}$

• Calculate the slope of $BH$ ($m_{BH}$): $m_{BH} = \frac{3 - 0}{4 - 0} = \frac{3}{4}$

• Apply the perpendicularity condition ($m_{BH} \cdot m_{AC} = -1$): Since $BH \perp AC$, we have: $m_{BH} \cdot m_{AC} = -1$ $\frac{3}{4} \cdot \frac{b - 2}{a} = -1$ $\frac{3(b - 2)}{4a} = -1$ $3(b - 2) = -4a$ $3b - 6 = -4a$ $4a + 3b = 6 \quad \ldots(2)$

Step 3: Solve the system of equations

We have two equations: (1) $4a + b = 0$ (2) $4a + 3b = 6$

Subtract equation (1) from equation (2): $(4a + 3b) - (4a + b) = 6 - 0$ $2b = 6$ $b = 3$

Substitute $b = 3$ into equation (1): $4a + 3 = 0$ $4a = -3$ $a = -\frac{3}{4}$

Therefore, the third vertex is $C\left(-\frac{3}{4}, 3\right)$.

Step 4: Determine the quadrant of the third vertex

The x-coordinate of the third vertex is $a = -\frac{3}{4}$, which is negative. The y-coordinate of the third vertex is $b = 3$, which is positive. A point with a negative x-coordinate and a positive y-coordinate lies in the second quadrant. However, the correct answer is given as third quadrant. Let's re-examine our steps.

The error lies in using BH perpendicular to AC instead of AH perpendicular to BC.

Step 2 (Revised): Use the fact that $AH$ is perpendicular to $BC$

The line segment $AH$ is an altitude and is therefore perpendicular to the side $BC$. We will find the slopes of $AH$ and $BC$, and then use the perpendicularity condition.

• Calculate the slope of $BC$ ($m_{BC}$): $m_{BC} = \frac{b - 3}{a - 4}$

• Calculate the slope of $AH$ ($m_{AH}$): $m_{AH} = \frac{2 - 0}{0 - 0}$ This is undefined, so $AH$ is a vertical line.

Since $AH$ is vertical, $BC$ must be horizontal. For $BC$ to be horizontal, we must have $b = 3$.

Substitute $b = 3$ into equation (1): $4a + 3 = 0$ $4a = -3$ $a = -\frac{3}{4}$

Therefore, the third vertex is $C\left(-\frac{3}{4}, 3\right)$. This is still in the second quadrant. There must be an error in the provided answer.

Lets verify the calculation with $BH \perp AC$ again

$m_{AC} = \frac{3-2}{-\frac{3}{4}-0} = \frac{1}{-\frac{3}{4}} = -\frac{4}{3}$ $m_{BH} = \frac{3-0}{4-0} = \frac{3}{4}$ $m_{AC} \cdot m_{BH} = -\frac{4}{3} \cdot \frac{3}{4} = -1$, so $BH \perp AC$ holds.

The solution is correct and the third vertex lies in the second quadrant. Since the given answer is the third quadrant, there must be an error.

Common Mistakes & Tips

• Always remember that the altitude is perpendicular to the side, not just a line extending from the vertex.
• Be careful with signs when calculating slopes and applying the perpendicularity condition.
• Double-check your algebra and arithmetic to avoid errors.
• If the orthocenter is the origin, using the origin to calculate slopes simplifies the calculations.

Summary

We used the properties of the orthocenter and altitudes of a triangle to set up a system of equations for the coordinates of the third vertex. By using the conditions that $CH \perp AB$ and $BH \perp AC$, we found the coordinates of the third vertex to be $(-\frac{3}{4}, 3)$, which lies in the second quadrant. The given correct answer of the third quadrant is incorrect.

Final Answer

The third vertex lies in the second quadrant. Since the question states that the correct answer is (A) third, there appears to be an error in the question or provided answer.