The othocenter of `DeltaABC` with vertices `B(1,-2)` and `C(-2,0)` is `H(3,-1)`.Find the vertex A.

Explanation

To find the vertex A of triangle ABC given the orthocenter H(3, -1), and the vertices B(1, -2) and C(-2, 0), we can follow these steps:

### Step 1: Identify the coordinates
Let the coordinates of vertex A be (x, y). We know:
- B = (1, -2)
- C = (-2, 0)
- H = (3, -1)

### Step 2: Find the slope of line BC
The slope of line BC can be calculated using the formula:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the coordinates of B and C:
\[ \text{slope of BC} = \frac{0 - (-2)}{-2 - 1} = \frac{2}{-3} = -\frac{2}{3} \]

### Step 3: Find the slope of line AH
Since AH is perpendicular to BC, the slope of AH is the negative reciprocal of the slope of BC:
\[ \text{slope of AH} = \frac{3}{2} \]

### Step 4: Write the equation of line AH
Using the point-slope form of the line equation \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is point H(3, -1):
\[ y - (-1) = \frac{3}{2}(x - 3) \]
This simplifies to:
\[ y + 1 = \frac{3}{2}x - \frac{9}{2} \]
Rearranging gives:
\[ y = \frac{3}{2}x - \frac{11}{2} \quad \text{(Equation 1)} \]

### Step 5: Find the slope of line BH
Now, we find the slope of line BH:
\[ \text{slope of BH} = \frac{-1 - (-2)}{3 - 1} = \frac{1}{2} \]

### Step 6: Find the slope of line AC
Since BH is perpendicular to AC, the slope of AC is the negative reciprocal of the slope of BH:
\[ \text{slope of AC} = -2 \]

### Step 7: Write the equation of line AC
Using point-slope form again with point B(1, -2):
\[ y - (-2) = -2(x - 1) \]
This simplifies to:
\[ y + 2 = -2x + 2 \]
Rearranging gives:
\[ y = -2x + 2 - 2 \Rightarrow y = -2x \quad \text{(Equation 2)} \]

### Step 8: Solve the equations simultaneously
Now we solve Equation 1 and Equation 2:
1. \(y = \frac{3}{2}x - \frac{11}{2}\)
2. \(y = -2x\)

Setting them equal:
\[ \frac{3}{2}x - \frac{11}{2} = -2x \]
Multiplying through by 2 to eliminate the fraction:
\[ 3x - 11 = -4x \]
Combining like terms:
\[ 3x + 4x = 11 \Rightarrow 7x = 11 \Rightarrow x = \frac{11}{7} \]

### Step 9: Substitute x back to find y
Substituting \(x = \frac{11}{7}\) into Equation 2:
\[ y = -2\left(\frac{11}{7}\right) = -\frac{22}{7} \]

### Final Answer
Thus, the coordinates of vertex A are:
\[ A\left(\frac{11}{7}, -\frac{22}{7}\right) \]