The orthocentre of the triangle whose vertices are (5,-2),(-1,2) and (1,4) is

Options

A. (1/5,14/5)
B. (14/5,1/5)
C. (1/5,1/5)
D. (14/5,14/5)

Answer

To find the orthocenter of the triangle with vertices A(5, -2), B(-1, 2), and C(1, 4), we will follow these steps:

### Step 1: Find the slopes of the sides of the triangle
1. **Calculate the slope of line BC:**
\[ \text{slope of } BC = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 2}{1 - (-1)} = \frac{2}{2} = 1 \]

2. **Calculate the slope of line AB:**
\[ \text{slope of } AB = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-2)}{-1 - 5} = \frac{4}{-6} = -\frac{2}{3} \]

### Step 2: Find the slopes of the altitudes
1. **The slope of altitude AD (perpendicular to BC):**
\[ \text{slope of } AD = -\frac{1}{\text{slope of } BC} = -\frac{1}{1} = -1 \]

2. **The slope of altitude CF (perpendicular to AB):**
\[ \text{slope of } CF = -\frac{1}{\text{slope of } AB} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2} \]

### Step 3: Write the equations of the altitudes
1. **Equation of altitude AD (passing through A(5, -2)):**
Using point-slope form \(y - y_1 = m(x - x_1)\):
\[ y + 2 = -1(x - 5) \implies y + 2 = -x + 5 \implies x + y = 3 \quad \text{(Equation 1)} \]

2. **Equation of altitude CF (passing through C(1, 4)):**
\[ y - 4 = \frac{3}{2}(x - 1) \implies y - 4 = \frac{3}{2}x - \frac{3}{2} \implies 3x - 2y = -5 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations to find the orthocenter
1. **Multiply Equation 1 by 3:**
\[ 3x + 3y = 9 \quad \text{(Equation 3)} \]

2. **Now, we have:**
- Equation 3: \(3x + 3y = 9\)
- Equation 2: \(3x - 2y = -5\)

3. **Subtract Equation 2 from Equation 3:**
\[ (3x + 3y) - (3x - 2y) = 9 - (-5) \implies 5y = 14 \implies y = \frac{14}{5} \]

4. **Substitute \(y = \frac{14}{5}\) into Equation 1:**
\[ x + \frac{14}{5} = 3 \implies x = 3 - \frac{14}{5} = \frac{15 - 14}{5} = \frac{1}{5} \]

### Final Result
The orthocenter of the triangle is at the point \(\left(\frac{1}{5}, \frac{14}{5}\right)\).

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