Question

The orthocentre of the triangle formed by lines \( x + y + 1 = 0 \), \( x - y - 1 = 0 \) and \( 3x + 4y + 5 = 0 \) is:

• A. \( (0,-1) \)
• B. \( (0,0) \)
• C. \( (1,1) \)
• D. \( (-1,0) \)

Hint:

The orthocentre of a triangle is the intersection of the altitudes. It can be found using perpendicular slopes and intersection points.

Answer 1

The Correct Option is A

Step 1: Finding the Intersection Points of the Given Lines To determine the orthocentre, we first find the vertices of the triangle by solving the equations in pairs. Solving \( x + y + 1 = 0 \) and \( x - y - 1 = 0 \): \[ x + y = -1 \] \[ x - y = 1 \] Adding both equations: \[ 2x = 0 \Rightarrow x = 0 \] Substituting \( x = 0 \) in \( x + y = -1 \): \[ y = -1 \] Thus, the intersection point is \( A(0,-1) \).
Step 2: Finding the Orthocentre The orthocentre of a triangle is the intersection of its altitudes. Given the symmetric structure of the lines, we find that the orthocentre lies at: \[ (0,-1) \]
Final Answer: \[ \boxed{(0,-1)} \]