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

To find the orthocenter of the triangle formed by the lines \(x + y + 1 = 0\), \(x - y - 1 = 0\), and \(3x + 4y + 5 = 0\), we first identify the vertices of the triangle by solving the equations of the lines in pairs:

1. Intersection of lines \(x+y+1=0\) and \(x-y-1=0\):

\(x+y+1=0\)
\(x-y-1=0\)

Adding the equations:
\(2x=0 \Rightarrow x=0\)
Substituting in \(x+y+1=0\):
\(0+y+1=0 \Rightarrow y=-1\)

Vertex A: \((0,-1)\)

2. Intersection of lines \(x-y-1=0\) and \(3x+4y+5=0\):

\(x-y-1=0\)
\(3x+4y+5=0\)

From \(x-y=1\), \(x=1+y\), substitute \(x=1+y\) into \(3x+4y+5=0\):
\(3(1+y)+4y+5=0\)
\(3+3y+4y+5=0\)
\(7y+8=0 \Rightarrow y=-\frac{8}{7}\)
Substituting back, \(x=1+(-\frac{8}{7})=\frac{-1}{7}\)

Vertex B: \((\frac{-1}{7},-\frac{8}{7})\)

3. Intersection of lines \(x+y+1=0\) and \(3x+4y+5=0\):

\(x+y+1=0\)
\(3x+4y+5=0\)

From \(x+y=-1\), express \(x=-1-y\), and substitute in:
\(3(-1-y)+4y+5=0\)
\(-3-3y+4y+5=0\)
\(y-2=0 \Rightarrow y=2\)
Substituting back, \(x=-1-2=-3\)

Vertex C: \((-3,2)\)

Having the vertices A \((0,-1)\), B \((\frac{-1}{7},-\frac{8}{7})\), and C \((-3,2)\), we determine the orthocenter by considering that it is the intersection of the altitudes. Since the line \(x-y-1=0\) is vertical and horizontal, the orthocenter lies on this axis:

The orthocenter is therefore located at the intersection of the identity lines:

Returning to vertex assumptions, the specific intersection deduced simplifies equally back to point \((0,-1)\) in context.

Therefore, the orthocenter is: (0,-1).

Answer 2

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: \[ \boxed{(0,-1)} \]