Question

Find the co-ordinates of the orthocentre of the triangle, the equations of whose sides are $x + y = 1$, $2x + 3y = 6$, $4x - y + 4 = 0$, without finding the co-ordinates of its vertices.

Answer 1

The coordinates of the orthocentre are $(\frac{3}{7}, \frac{22}{7})$.

Answer 2

To find the orthocentre of the triangle without finding the coordinates of its vertices, we use the property that an altitude passes through a vertex and is perpendicular to the opposite side. We also use the concept of a line passing through the intersection of two lines.

Let the equations of the sides of the triangle be:

1. $L_1: x + y - 1 = 0$
2. $L_2: 2x + 3y - 6 = 0$
3. $L_3: 4x - y + 4 = 0$

Let the vertices be A, B, C.

• Vertex A is the intersection of $L_2$ and $L_3$.
• Vertex B is the intersection of $L_1$ and $L_3$.
• Vertex C is the intersection of $L_1$ and $L_2$.

The orthocentre is the intersection of the altitudes. We need to find the equations of at least two altitudes.

Step 1: Find the equation of the altitude from vertex A to side $L_1$.

The altitude from A passes through the intersection of $L_2$ and $L_3$. Its equation can be written in the form $L_2 + \lambda L_3 = 0$.

$(2x + 3y - 6) + \lambda (4x - y + 4) = 0$

$(2 + 4\lambda)x + (3 - \lambda)y + (-6 + 4\lambda) = 0$

This altitude is perpendicular to side $L_1: x + y - 1 = 0$.

The slope of $L_1$ is $m_1 = -1/1 = -1$.

The slope of the altitude must be $m_{alt1} = -1/m_1 = -1/(-1) = 1$.

The slope of the altitude $(2 + 4\lambda)x + (3 - \lambda)y + (-6 + 4\lambda) = 0$ is $m_{alt1} = -\frac{2 + 4\lambda}{3 - \lambda}$.

Equating the slopes:

$-\frac{2 + 4\lambda}{3 - \lambda} = 1$

$-(2 + 4\lambda) = 3 - \lambda$

$-2 - 4\lambda = 3 - \lambda$

$-5 = 3\lambda \Rightarrow \lambda = -5/3$

Substitute $\lambda = -5/3$ into the altitude equation:

$(2 + 4(-5/3))x + (3 - (-5/3))y + (-6 + 4(-5/3)) = 0$

$(2 - 20/3)x + (3 + 5/3)y + (-6 - 20/3) = 0$

$(-14/3)x + (14/3)y + (-38/3) = 0$

Multiply by $-3/14$:

$x - y + 38/14 = 0$

$x - y + 19/7 = 0$

Multiplying by 7, we get the first altitude equation:

$A_1: 7x - 7y + 19 = 0$

Step 2: Find the equation of the altitude from vertex B to side $L_2$.

The altitude from B passes through the intersection of $L_1$ and $L_3$. Its equation can be written in the form $L_1 + \mu L_3 = 0$.

$(x + y - 1) + \mu (4x - y + 4) = 0$

$(1 + 4\mu)x + (1 - \mu)y + (-1 + 4\mu) = 0$

This altitude is perpendicular to side $L_2: 2x + 3y - 6 = 0$.

The slope of $L_2$ is $m_2 = -2/3$.

The slope of the altitude must be $m_{alt2} = -1/m_2 = -1/(-2/3) = 3/2$.

The slope of the altitude $(1 + 4\mu)x + (1 - \mu)y + (-1 + 4\mu) = 0$ is $m_{alt2} = -\frac{1 + 4\mu}{1 - \mu}$.

Equating the slopes:

$-\frac{1 + 4\mu}{1 - \mu} = 3/2$

$-2(1 + 4\mu) = 3(1 - \mu)$

$-2 - 8\mu = 3 - 3\mu$

$-5 = 5\mu \Rightarrow \mu = -1$

Substitute $\mu = -1$ into the altitude equation:

$(1 + 4(-1))x + (1 - (-1))y + (-1 + 4(-1)) = 0$

$(1 - 4)x + (1 + 1)y + (-1 - 4) = 0$

$-3x + 2y - 5 = 0$

Multiplying by -1, we get the second altitude equation:

$A_2: 3x - 2y + 5 = 0$

Step 3: Find the intersection of the two altitudes to get the orthocentre.

We need to solve the system of equations for $A_1$ and $A_2$:

1. $7x - 7y + 19 = 0 \Rightarrow 7x - 7y = -19$
2. $3x - 2y + 5 = 0 \Rightarrow 3x - 2y = -5$

Multiply equation (1) by 2 and equation (2) by 7 to eliminate y:

$2 \times (7x - 7y = -19) \Rightarrow 14x - 14y = -38$

$7 \times (3x - 2y = -5) \Rightarrow 21x - 14y = -35$

Subtract the first new equation from the second new equation:

$(21x - 14y) - (14x - 14y) = -35 - (-38)$

$7x = 3$

$x = 3/7$

Substitute $x = 3/7$ into equation (2):

$3(3/7) - 2y = -5$

$9/7 - 2y = -5$

$-2y = -5 - 9/7$

$-2y = (-35 - 9)/7$

$-2y = -44/7$

$y = 22/7$

Thus, the coordinates of the orthocentre are $(3/7, 22/7)$.