The orthocentre of the triangle formed by the lines `x=2,y=3` and `3x+2y=6` is at the point

Options

A. (2,0)
B. (0,3)
C. (2,3)
D. none of these

Explanation

To find the orthocenter of the triangle formed by the lines \( x = 2 \), \( y = 3 \), and \( 3x + 2y = 6 \), we will follow these steps:

### Step 1: Identify the vertices of the triangle
The vertices of the triangle can be found by determining the points of intersection of the given lines.

1. **Intersection of \( x = 2 \) and \( y = 3 \)**:
- This gives us the point \( A(2, 3) \).

2. **Intersection of \( x = 2 \) and \( 3x + 2y = 6 \)**:
- Substitute \( x = 2 \) into the equation:
\[ 3(2) + 2y = 6 \implies 6 + 2y = 6 \implies 2y = 0 \implies y = 0 \]
- This gives us the point \( B(2, 0) \).

3. **Intersection of \( y = 3 \) and \( 3x + 2y = 6 \)**:
- Substitute \( y = 3 \) into the equation:
\[ 3x + 2(3) = 6 \implies 3x + 6 = 6 \implies 3x = 0 \implies x = 0 \]
- This gives us the point \( C(0, 3) \).

### Step 2: Determine the slopes of the sides
Next, we will find the slopes of the sides of the triangle:

1. **Slope of line \( AB \)** (between points \( A(2, 3) \) and \( B(2, 0) \)):
- Since both points have the same \( x \)-coordinate, the slope is undefined (vertical line).

2. **Slope of line \( BC \)** (between points \( B(2, 0) \) and \( C(0, 3) \)):
- The slope \( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{0 - 2} = \frac{3}{-2} = -\frac{3}{2} \).

3. **Slope of line \( AC \)** (between points \( A(2, 3) \) and \( C(0, 3) \)):
- Since both points have the same \( y \)-coordinate, the slope is 0 (horizontal line).

### Step 3: Determine the altitudes
Since \( AB \) is vertical and \( AC \) is horizontal, we can conclude that:

- The altitude from \( C \) (perpendicular to \( AB \)) is the horizontal line through \( C \) (which is \( y = 3 \)).
- The altitude from \( A \) (perpendicular to \( BC \)) can be found using the negative reciprocal of the slope of \( BC \):
- The slope of \( BC \) is \( -\frac{3}{2} \), so the slope of the altitude from \( A \) is \( \frac{2}{3} \).
- The equation of the line through \( A(2, 3) \) with slope \( \frac{2}{3} \) is:
\[ y - 3 = \frac{2}{3}(x - 2) \implies y = \frac{2}{3}x + \frac{5}{3} \]

### Step 4: Find the orthocenter
The orthocenter is the intersection of the altitudes. We need to find the intersection of the line \( y = 3 \) (altitude from \( C \)) and the line \( y = \frac{2}{3}x + \frac{5}{3} \) (altitude from \( A \)).

1. Set \( y = 3 \) equal to \( \frac{2}{3}x + \frac{5}{3} \):
\[ 3 = \frac{2}{3}x + \frac{5}{3} \]
Multiply through by 3 to eliminate the fraction:
\[ 9 = 2x + 5 \implies 2x = 4 \implies x = 2 \]

2. Substitute \( x = 2 \) back into \( y = 3 \):
- Thus, the orthocenter is at the point \( (2, 3) \).

### Conclusion
The orthocenter of the triangle formed by the lines \( x = 2 \), \( y = 3 \), and \( 3x + 2y = 6 \) is at the point \( (2, 3) \).

---