Key Concepts and Formulas

• The orthocentre of a triangle is the point of intersection of its altitudes. In a right-angled triangle, the orthocentre is the vertex at the right angle.
• The slope of a line given by the equation $Ax + By + C = 0$ is $m = -\frac{A}{B}$. Two lines are perpendicular if the product of their slopes is $-1$.
• The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Step-by-Step Solution

We need to find the distance between the orthocentres of two triangles. Let's find each orthocentre.

Part 1: Finding the Orthocentre of the First Triangle

The first triangle is formed by the lines: $L_1: 4x - 7y + 10 = 0$ $L_2: x + y = 5$ $L_3: 7x + 4y = 15$

Step 1: Calculate the slopes of the lines to check for perpendicularity. We calculate the slopes of the lines using the formula $m = -\frac{A}{B}$.

• Slope of $L_1$: $m_1 = -\frac{4}{-7} = \frac{4}{7}$
• Slope of $L_2$: $m_2 = -\frac{1}{1} = -1$
• Slope of $L_3$: $m_3 = -\frac{7}{4}$

Step 2: Check if the triangle is right-angled. We check if the product of any two slopes is $-1$.

• $m_1 \cdot m_2 = (\frac{4}{7}) \cdot (-1) = -\frac{4}{7} \neq -1$
• $m_2 \cdot m_3 = (-1) \cdot (-\frac{7}{4}) = \frac{7}{4} \neq -1$
• $m_1 \cdot m_3 = (\frac{4}{7}) \cdot (-\frac{7}{4}) = -1$

Since $m_1 \cdot m_3 = -1$, lines $L_1$ and $L_3$ are perpendicular. The triangle formed by these three lines is a right-angled triangle.

Step 3: Determine the orthocentre of the right-angled triangle. The orthocentre of a right-angled triangle is the vertex where the right angle is formed, which is the intersection of $L_1$ and $L_3$. We solve the system of equations: $L_1: 4x - 7y = -10$ $L_3: 7x + 4y = 15$

Multiply the first equation by 4 and the second by 7 to eliminate $y$: $16x - 28y = -40$ $49x + 28y = 105$

Adding the two equations gives: $65x = 65$ $x = 1$

Substitute $x = 1$ into the first equation: $4(1) - 7y = -10$ $4 - 7y = -10$ $-7y = -14$ $y = 2$

The intersection point is $(1, 2)$. Therefore, the orthocentre of the first triangle, $H_1$, is $(1, 2)$.

Part 2: Finding the Orthocentre of the Second Triangle

The second triangle is formed by the lines: $L_a: x = 0$ (y-axis) $L_b: y = 0$ (x-axis) $L_c: x + y = 1$

Step 1: Calculate the slopes of the lines.

• Slope of $L_a$ ($x = 0$): Undefined (vertical line)
• Slope of $L_b$ ($y = 0$): $0$ (horizontal line)
• Slope of $L_c$ ($x + y = 1$): $-1$

Step 2: Identify if the triangle is right-angled. Since $x = 0$ and $y = 0$ are perpendicular, the triangle is right-angled.

Step 3: Determine the orthocentre of the right-angled triangle. The right angle is formed by the intersection of $x = 0$ and $y = 0$, which is the origin $(0, 0)$. Therefore, the orthocentre of the second triangle, $H_2$, is $(0, 0)$.

Part 3: Calculating the Distance Between the Two Orthocentres

We have $H_1 = (1, 2)$ and $H_2 = (0, 0)$.

Step 1: Use the distance formula. $D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ $D = \sqrt{(0 - 1)^2 + (0 - 2)^2}$ $D = \sqrt{(-1)^2 + (-2)^2}$ $D = \sqrt{1 + 4}$ $D = \sqrt{5}$

The distance between the orthocentres is $\sqrt{5}$.

Common Mistakes & Tips

• Always check for right-angled triangles first to simplify the problem.
• Remember the properties of orthocentres, circumcentres, and incentres.
• Be careful with signs when calculating slopes and distances.

Summary

The problem involves finding the distance between the orthocentres of two triangles. By recognizing that both triangles are right-angled, we simplified the process of finding the orthocentres. The first orthocentre is $(1, 2)$ and the second is $(0, 0)$. The distance between them is $\sqrt{5}$.

The final answer is $\boxed{\sqrt{5}}$, which corresponds to option (C).