Question

The orthocenter of the triangle whose sides are given by x + y + 10 = 0, x - y - 2 = 0 and 2x + y - 7 = 0 is

• A. (-4, -3)
• B. (-4, -6)
• C. (4,6)
• D. (3,6)

Answer 1

The Correct Option is B

To find the vertex of the triangle formed by the lines \(x + y + 10 = 0\), \(x - y - 2 = 0\), and \(2x + y - 7 = 0\)

1. Finding the First Vertex:
Solve for the intersection of \(x + y + 10 = 0\) and \(x - y - 2 = 0\):
\( x + y = -10 \quad (1) \)
\( x - y = 2 \quad (2) \)
Add the equations:

\( (x + y) + (x - y) = -10 + 2 \)
\( 2x = -8 \implies x = -4 \)
Substitute \(x = -4\) into (2):

\( -4 - y = 2 \implies y = -6 \)
Thus, the intersection is \((-4, -6)\).

2. Verifying Other Vertices:
Solve for the intersection of \(x - y - 2 = 0\) and \(2x + y - 7 = 0\):

\( x - y = 2 \quad (1) \)
\( 2x + y = 7 \quad (2) \)
Add the equations:

\( (x - y) + (2x + y) = 2 + 7 \)
\( 3x = 9 \implies x = 3 \)
Substitute \(x = 3\) into (1):

\( 3 - y = 2 \implies y = 1 \)
So, the intersection is \((3, 1)\).

3. Third Vertex:
Solve for the intersection of \(x + y + 10 = 0\) and \(2x + y - 7 = 0\):

\( x + y = -10 \quad (1) \)
\( 2x + y = 7 \quad (2) \)
Subtract (1) from (2):

\( (2x + y) - (x + y) = 7 - (-10) \)
\( x = 17 \)
Substitute \(x = 17\) into (1):

\( 17 + y = -10 \implies y = -27 \)
So, the intersection is \((17, -27)\).

\( 2(-4) - 3(-6) + k = -8 + 18 + k = 10 + k = 0 \implies k = -10 \)
However, using \(2x - 3y - 10 = 0\) makes all three lines concurrent at \((-4, -6)\), which does not form a triangle. 
Thus, the original line \(2x + y - 7 = 0\) is likely correct, as it yields \((-4, -6)\) as a vertex.

Final Answer:
The vertex is \((-4, -6)\).