The orthocentre of the triangle formed by the points (0, 0), (8, 0) and (4, 6) is

Options

A. `(4,(8)/(3))`
B. (3,4)
C. (4,3)
D. `(3,(5)/(2))`

Answer

To find the orthocenter of the triangle formed by the points (0, 0), (8, 0), and (4, 6), we will follow these steps:

### Step 1: Identify the vertices of the triangle
Let the vertices of the triangle be:
- A(0, 0)
- B(8, 0)
- C(4, 6)

### Step 2: Calculate the slopes of the sides of the triangle
1. **Slope of side BC**:
\[ \text{slope of BC} = \frac{y_C - y_B}{x_C - x_B} = \frac{6 - 0}{4 - 8} = \frac{6}{-4} = -\frac{3}{2} \]

2. **Slope of side CA**:
\[ \text{slope of CA} = \frac{y_A - y_C}{x_A - x_C} = \frac{0 - 6}{0 - 4} = \frac{-6}{-4} = \frac{3}{2} \]

### Step 3: Find the slopes of the altitudes
1. **Slope of altitude from A to BC** (perpendicular to BC):
\[ \text{slope} = -\frac{1}{\text{slope of BC}} = -\frac{1}{-\frac{3}{2}} = \frac{2}{3} \]

2. **Slope of altitude from B to CA** (perpendicular to CA):
\[ \text{slope} = -\frac{1}{\text{slope of CA}} = -\frac{1}{\frac{3}{2}} = -\frac{2}{3} \]

### Step 4: Write the equations of the altitudes
1. **Equation of altitude from A (0, 0) to BC**:
Using point-slope form \(y - y_1 = m(x - x_1)\):
\[ y - 0 = \frac{2}{3}(x - 0) \implies y = \frac{2}{3}x \]

2. **Equation of altitude from B (8, 0) to CA**:
\[ y - 0 = -\frac{2}{3}(x - 8) \implies y = -\frac{2}{3}x + \frac{16}{3} \]

### Step 5: Solve the equations to find the orthocenter
We have the two equations:
1. \(y = \frac{2}{3}x\)
2. \(y = -\frac{2}{3}x + \frac{16}{3}\)

Setting them equal to each other:
\[ \frac{2}{3}x = -\frac{2}{3}x + \frac{16}{3} \]

Combining like terms:
\[ \frac{2}{3}x + \frac{2}{3}x = \frac{16}{3} \]
\[ \frac{4}{3}x = \frac{16}{3} \]

Multiplying both sides by \(\frac{3}{4}\):
\[ x = 4 \]

Now substituting \(x = 4\) back into the first equation to find \(y\):
\[ y = \frac{2}{3}(4) = \frac{8}{3} \]

### Conclusion
The orthocenter of the triangle is at the point \((4, \frac{8}{3})\).

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