Key Concepts and Formulas

• Orthocentre: The point of intersection of the altitudes of a triangle.
• Slope of a line: Given two points $(x_1, y_1)$ and $(x_2, y_2)$, the slope is $m = \frac{y_2 - y_1}{x_2 - x_1}$.
• Perpendicular lines: Two lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $m_1 m_2 = -1$.
• Quadratic equation: A quadratic equation with roots $r_1$ and $r_2$ can be written as $x^2 - (r_1 + r_2)x + r_1 r_2 = 0$.

Step-by-Step Solution

Step 1: Define the Vertices and Orthocentre

Let the vertices of the triangle be $A(1, 2)$, $B(2, 3)$, and $C(3, 1)$. We are given that the orthocentre is $H(\alpha, \beta)$. Our goal is to find $\alpha$ and $\beta$.

Step 2: Calculate Slopes of the Sides of the Triangle

To find the equations of the altitudes, we first need the slopes of the sides to which they are perpendicular.

• Slope of side AC ($m_{AC}$): Using points $A(1, 2)$ and $C(3, 1)$: $m_{AC} = \frac{1-2}{3-1} = \frac{-1}{2} = -\frac{1}{2}$

• Slope of side BC ($m_{BC}$): Using points $B(2, 3)$ and $C(3, 1)$: $m_{BC} = \frac{1-3}{3-2} = \frac{-2}{1} = -2$

Step 3: Formulate Equations for the Altitudes

The orthocentre is the intersection of the altitudes. We only need two altitudes to find their intersection point.

• Altitude from B to AC (let's call it $L_B$): This altitude passes through vertex $B(2, 3)$. It is perpendicular to side AC. Since $L_B \perp AC$, the product of their slopes must be -1: $m_{L_B} \times m_{AC} = -1$. Therefore, $m_{L_B} = -\frac{1}{m_{AC}} = -\frac{1}{-\frac{1}{2}} = 2$. The equation of the altitude $L_B$ passing through $B(2,3)$ with slope $2$ is given by $y - 3 = 2(x - 2)$, which simplifies to $y = 2x - 4 + 3$, or $y = 2x - 1$. Since the orthocentre $H(\alpha, \beta)$ lies on this line, we have: $\beta = 2\alpha - 1 \quad \text{(Equation 1)}$

• Altitude from A to BC (let's call it $L_A$): This altitude passes through vertex $A(1, 2)$. It is perpendicular to side BC. Since $L_A \perp BC$, the product of their slopes must be -1: $m_{L_A} \times m_{BC} = -1$. Therefore, $m_{L_A} = -\frac{1}{m_{BC}} = -\frac{1}{-2} = \frac{1}{2}$. The equation of the altitude $L_A$ passing through $A(1,2)$ with slope $\frac{1}{2}$ is given by $y - 2 = \frac{1}{2}(x - 1)$, which simplifies to $2y - 4 = x - 1$, or $2y = x + 3$. Since the orthocentre $H(\alpha, \beta)$ lies on this line, we have: $2\beta = \alpha + 3 \quad \text{(Equation 2)}$

Step 4: Solve the System of Equations for $\alpha$ and $\beta$

We now have a system of two linear equations with two variables:

1. $\beta = 2\alpha-1$
2. $2\beta = \alpha+3$

Substitute Equation 1 into Equation 2: $2(2\alpha-1) = \alpha+3$ $4\alpha-2 = \alpha+3$ $3\alpha = 5$ $\alpha = \frac{5}{3}$

Now substitute the value of $\alpha$ back into Equation 1 to find $\beta$: $\beta = 2\left(\frac{5}{3}\right)-1$ $\beta = \frac{10}{3}-1$ $\beta = \frac{7}{3}$

So, the orthocentre of the triangle is $H\left(\frac{5}{3}, \frac{7}{3}\right)$.

Step 5: Calculate the Roots of the Quadratic Equation

The problem asks for a quadratic equation whose roots are $\alpha+4\beta$ and $4\alpha+\beta$.

Let the first root be $r_1 = \alpha+4\beta$: $r_1 = \frac{5}{3} + 4\left(\frac{7}{3}\right) = \frac{5}{3} + \frac{28}{3} = \frac{33}{3} = 11$

Let the second root be $r_2 = 4\alpha+\beta$: $r_2 = 4\left(\frac{5}{3}\right) + \frac{7}{3} = \frac{20}{3} + \frac{7}{3} = \frac{27}{3} = 9$

Step 6: Form the Quadratic Equation

A quadratic equation with roots $r_1$ and $r_2$ is given by the formula: $x^2 - (r_1+r_2)x + r_1r_2 = 0$

Calculate the sum of the roots ($S$): $S = r_1+r_2 = 11+9 = 20$

Calculate the product of the roots ($P$): $P = r_1r_2 = 11 \times 9 = 99$

Substitute these values into the quadratic formula: $x^2 - (20)x + (99) = 0$ $x^2 - 20x + 99 = 0$

Common Mistakes & Tips:

• Sign Errors: Be extremely careful with signs when calculating slopes and substituting values. A small sign error can lead to a completely wrong answer.
• Equation of a Line: Remember the point-slope form of a line: $y - y_1 = m(x - x_1)$.
• System of Equations: Double-check your algebraic manipulations when solving the system of equations to avoid errors.

Summary:

We found the orthocentre of the triangle by finding the intersection of two altitudes. This involved calculating the slopes of the triangle's sides and using the perpendicularity condition to find the equations of the altitudes. Solving the resulting system of linear equations gave us the coordinates of the orthocentre $(\alpha, \beta)$. Finally, we used these values to compute the two roots, $\alpha+4\beta$ and $4\alpha+\beta$, and constructed the quadratic equation $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.

The final answer is \boxed{x^2-20x+99=0}, which corresponds to option (A).