If the orthocentre of the triangle,whose vertices are `(1,2),(2,3)` and `(3,1)` is `(alpha,beta)`,then the quadratic equation whose roots are `alpha+4 beta` and `4 alpha+beta`,is

Options

A. `x^(2)-19x+90=0`
B. `x^(2)-20x+99=0`
C. `x^(2)-18x+80=0`
D. `x^(2)-22x+120=0`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Find the coordinates of the orthocenter of the triangle with vertices A(1, 2), B(2, 3), and C(3, 1).

The orthocenter of a triangle can be found using the intersection of the altitudes. We first need to find the slopes of the sides of the triangle.

1. **Find the slope of AB:**
\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 2}{2 - 1} = \frac{1}{1} = 1 \]

2. **Find the slope of BC:**
\[ m_{BC} = \frac{y_3 - y_2}{x_3 - x_2} = \frac{1 - 3}{3 - 2} = \frac{-2}{1} = -2 \]

3. **Find the slope of AC:**
\[ m_{AC} = \frac{y_3 - y_1}{x_3 - x_1} = \frac{1 - 2}{3 - 1} = \frac{-1}{2} \]

### Step 2: Find the equations of the altitudes.

1. **Altitude from C (perpendicular to AB):**
The slope of the altitude from C is the negative reciprocal of the slope of AB:
\[ m_{CE} = -\frac{1}{m_{AB}} = -1 \]
Using point C(3, 1):
\[ y - 1 = -1(x - 3) \implies y = -x + 4 \quad \text{(Equation 1)} \]

2. **Altitude from A (perpendicular to BC):**
The slope of the altitude from A is the negative reciprocal of the slope of BC:
\[ m_{AD} = -\frac{1}{m_{BC}} = \frac{1}{2} \]
Using point A(1, 2):
\[ y - 2 = \frac{1}{2}(x - 1) \implies 2y - 4 = x - 1 \implies x - 2y + 3 = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations to find the orthocenter (α, β).

Now we solve the two equations (Equation 1 and Equation 2) simultaneously:
1. From Equation 1:
\[ y = -x + 4 \]
Substitute into Equation 2:
\[ x - 2(-x + 4) + 3 = 0 \implies x + 2x - 8 + 3 = 0 \implies 3x - 5 = 0 \implies x = \frac{5}{3} \]

2. Substitute \( x = \frac{5}{3} \) back into Equation 1:
\[ y = -\frac{5}{3} + 4 = -\frac{5}{3} + \frac{12}{3} = \frac{7}{3} \]

Thus, the orthocenter is:
\[ (\alpha, \beta) = \left(\frac{5}{3}, \frac{7}{3}\right) \]

### Step 4: Find the roots for the quadratic equation.

The roots are given as:
1. \( \alpha + 4\beta = \frac{5}{3} + 4 \times \frac{7}{3} = \frac{5 + 28}{3} = \frac{33}{3} = 11 \)
2. \( 4\alpha + \beta = 4 \times \frac{5}{3} + \frac{7}{3} = \frac{20 + 7}{3} = \frac{27}{3} = 9 \)

### Step 5: Form the quadratic equation.

The sum of the roots \( S = 11 + 9 = 20 \) and the product of the roots \( P = 11 \times 9 = 99 \).

The quadratic equation is:
\[ x^2 - Sx + P = 0 \implies x^2 - 20x + 99 = 0 \]

### Final Answer:
The quadratic equation is:
\[ \boxed{x^2 - 20x + 99 = 0} \]