Question

If the orthocentre of the triangle formed by the lines $ y = x + 1 $, $ y = 4x - 8 $, and $ y = mx + c $ is at $ (3, -1) $, then $ m - c $ is:

• A. 0
• B. -2
• C. 4
• D. 2

Hint:

When finding the orthocenter, use the condition that the perpendiculars from the vertices to the opposite sides intersect at the orthocenter, and solve for the unknowns step by step.

Answer 1

The Correct Option is A

Step 1: Find the equations of the lines
We are given three lines:
1. \( y = x + 1 \) (Line 1) 2. \( y = 4x - 8 \) (Line 2) 3. \( y = mx + c \) (Line 3) 
Let the orthocenter of the triangle formed by these three lines be at the point \( (3, -1) \). 
Step 2: Find the intersection points of the lines
Intersection of Line 1 and Line 2: The equations of Line 1 and Line 2 are:
\[ y = x + 1 \quad \text{and} \quad y = 4x - 8 \] Equating the two equations: \[ x + 1 = 4x - 8 \] 
Solving for \( x \): \[ x = 3 \] Substitute \( x = 3 \) into \( y = x + 1 \) to find \( y \): \[ y = 3 + 1 = 4 \] 
Thus, the point of intersection of Line 1 and Line 2 is \( (3, 4) \). 
- Intersection of Line 1 and Line 3: The equations of Line 1 and Line 3 are: \[ y = x + 1 \quad \text{and} \quad y = mx + c \] Equating the two equations: \[ x + 1 = mx + c \] Rearranging: \[ x(1 - m) = c - 1 \] \[ x = \frac{c - 1}{1 - m} \] Substitute \( x = \frac{c - 1}{1 - m} \) and \( y = x + 1 \) into the equation for the orthocenter to find \( m - c \). 
Step 3: Using the condition that the orthocenter is at \( (3, -1) \)
Given that the orthocenter is at the point \( (3, -1) \), substitute this point into the equations and solve for \( m \) and \( c \). 
After performing the calculations, we find that \( m - c = 0 \).