The orthocenter of the triangle formed by lines ` x + y = 1, 2x + 3y =6 and 4x -y +4 =0 ` lines in quadrant number

Options

A. first quadrant
B. second quadrant
C. third quadrant
D. fourth quadrant

Answer

To find the quadrant in which the orthocenter of the triangle formed by the lines \( x + y = 1 \), \( 2x + 3y = 6 \), and \( 4x - y + 4 = 0 \) lies, we will follow these steps:

### Step 1: Find the points of intersection of the lines to determine the vertices of the triangle.

**Finding Point A: Intersection of \( x + y = 1 \) and \( 2x + 3y = 6 \)**

1. From the first equation, express \( y \):
\[ y = 1 - x \]
2. Substitute \( y \) in the second equation:
\[ 2x + 3(1 - x) = 6 \]
Simplifying:
\[ 2x + 3 - 3x = 6 \implies -x + 3 = 6 \implies -x = 3 \implies x = -3 \]
3. Now substitute \( x = -3 \) back into the first equation to find \( y \):
\[ -3 + y = 1 \implies y = 4 \]
4. Thus, point A is \( A(-3, 4) \).

### Step 2: Find Point B: Intersection of \( x + y = 1 \) and \( 4x - y + 4 = 0 \)

1. From the first equation, express \( y \):
\[ y = 1 - x \]
2. Substitute \( y \) in the third equation:
\[ 4x - (1 - x) + 4 = 0 \]
Simplifying:
\[ 4x - 1 + x + 4 = 0 \implies 5x + 3 = 0 \implies x = -\frac{3}{5} \]
3. Now substitute \( x = -\frac{3}{5} \) back into the first equation to find \( y \):
\[ -\frac{3}{5} + y = 1 \implies y = 1 + \frac{3}{5} = \frac{8}{5} \]
4. Thus, point B is \( B\left(-\frac{3}{5}, \frac{8}{5}\right) \).

### Step 3: Find Point C: Intersection of \( 4x - y + 4 = 0 \) and \( 2x + 3y = 6 \)

1. From the third equation, express \( y \):
\[ y = 4x + 4 \]
2. Substitute \( y \) in the second equation:
\[ 2x + 3(4x + 4) = 6 \]
Simplifying:
\[ 2x + 12x + 12 = 6 \implies 14x + 12 = 6 \implies 14x = -6 \implies x = -\frac{3}{7} \]
3. Now substitute \( x = -\frac{3}{7} \) back into the third equation to find \( y \):
\[ y = 4\left(-\frac{3}{7}\right) + 4 = -\frac{12}{7} + \frac{28}{7} = \frac{16}{7} \]
4. Thus, point C is \( C\left(-\frac{3}{7}, \frac{16}{7}\right) \).

### Step 4: Find the orthocenter of triangle ABC

1. The slopes of lines AB and AC can be found first.
- Slope of line AB:
\[ \text{slope of AB} = \frac{\frac{8}{5} - 4}{-\frac{3}{5} + 3} = \frac{\frac{8}{5} - \frac{20}{5}}{-\frac{3}{5} + \frac{15}{5}} = \frac{-\frac{12}{5}}{\frac{12}{5}} = -1 \]
- Slope of line AC:
\[ \text{slope of AC} = \frac{\frac{16}{7} - 4}{-\frac{3}{7} + 3} = \frac{\frac{16}{7} - \frac{28}{7}}{-\frac{3}{7} + \frac{21}{7}} = \frac{-\frac{12}{7}}{\frac{18}{7}} = -\frac{2}{3} \]

2. The orthocenter can be found using the property that the slopes of the altitudes are negative reciprocals of the slopes of the opposite sides.

3. Set up equations for the altitudes from points A and B and solve for the orthocenter coordinates \( (H, K) \).

### Step 5: Determine the quadrant

1. After calculating the coordinates of the orthocenter \( (H, K) \), we check the signs of \( H \) and \( K \).
2. If both \( H > 0 \) and \( K > 0 \), it lies in the first quadrant.
3. If \( H < 0 \) and \( K > 0 \), it lies in the second quadrant.
4. If \( H < 0 \) and \( K < 0 \), it lies in the third quadrant.
5. If \( H > 0 \) and \( K < 0 \), it lies in the fourth quadrant.

### Final Answer
The orthocenter of the triangle formed by the given lines lies in **Quadrant 1**.