Find the orthocentre of the triangle whose sides have the equations y = 15, 3x = 4y, and 5x + 12y = 0.

Options

A. (0, 0)
B. (0, -33)
C. (- 33, 33)
D. (- 33, 0)

Answer

To find the orthocenter of the triangle formed by the lines given by the equations \(y = 15\), \(3x = 4y\), and \(5x + 12y = 0\), we will follow these steps:

### Step 1: Find the points of intersection of the lines

1. **Intersection of \(y = 15\) and \(3x = 4y\)**:
- Substitute \(y = 15\) into \(3x = 4y\):
\[ 3x = 4(15) \implies 3x = 60 \implies x = 20 \]
- Thus, the point of intersection is \(A(20, 15)\).

2. **Intersection of \(y = 15\) and \(5x + 12y = 0\)**:
- Substitute \(y = 15\) into \(5x + 12y = 0\):
\[ 5x + 12(15) = 0 \implies 5x + 180 = 0 \implies 5x = -180 \implies x = -36 \]
- Thus, the point of intersection is \(B(-36, 15)\).

3. **Intersection of \(3x = 4y\) and \(5x + 12y = 0\)**:
- Rewrite \(3x = 4y\) as \(y = \frac{3}{4}x\).
- Substitute \(y = \frac{3}{4}x\) into \(5x + 12y = 0\):
\[ 5x + 12\left(\frac{3}{4}x\right) = 0 \implies 5x + 9x = 0 \implies 14x = 0 \implies x = 0 \]
- Substitute \(x = 0\) back into \(y = \frac{3}{4}x\):
\[ y = \frac{3}{4}(0) = 0 \]
- Thus, the point of intersection is \(O(0, 0)\).

### Step 2: Determine the equations of the altitudes

1. **Altitude from point \(O(0, 0)\) to line \(AB\)** (which is horizontal, \(y = 15\)):
- The altitude will be a vertical line through \(O\), which is \(x = 0\).

2. **Altitude from point \(B(-36, 15)\) to line \(AC\)**:
- The slope of line \(AC\) (from \(A(20, 15)\) to \(O(0, 0)\)) is:
\[ \text{slope} = \frac{15 - 0}{20 - 0} = \frac{15}{20} = \frac{3}{4} \]
- The slope of the altitude from \(B\) will be the negative reciprocal:
\[ \text{slope} = -\frac{4}{3} \]
- Using point-slope form \(y - y_1 = m(x - x_1)\):
\[ y - 15 = -\frac{4}{3}(x + 36) \]
- Simplifying:
\[ y - 15 = -\frac{4}{3}x - 48 \implies y = -\frac{4}{3}x - 33 \]

### Step 3: Find the orthocenter by solving the equations of the altitudes

- We have the equations:
1. \(x = 0\) (altitude from \(O\))
2. \(y = -\frac{4}{3}x - 33\) (altitude from \(B\))

- Substitute \(x = 0\) into the second equation:
\[ y = -\frac{4}{3}(0) - 33 = -33 \]

### Conclusion

Thus, the orthocenter of the triangle is at the point \(H(0, -33)\).