The orthocenter of the triangle whose sides are given by x + y + 10 = 0, x - y - 2 = 0 and 2x + y - 7 = 0 is
The orthocenter of the triangle whose sides are given by x + y + 10 = 0, x - y - 2 = 0 and 2x + y - 7 = 0 is
(-4, -3)
(-4, -6)
(4,6)
(3,6)
The Correct Option is B
Solution and Explanation
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 \((-4, -6)\).
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Concepts Used:
Three Dimensional Geometry
Mathematically, Geometry is one of the most important topics. The concepts of Geometry are derived w.r.t. the planes. So, Geometry is divided into three major categories based on its dimensions which are one-dimensional geometry, two-dimensional geometry, and three-dimensional geometry.
Direction Cosines and Direction Ratios of Line:
Consider a line L that is passing through the three-dimensional plane. Now, x,y and z are the axes of the plane and α,β, and γ are the three angles the line makes with these axes. These are commonly known as the direction angles of the plane. So, appropriately, we can say that cosα, cosβ, and cosγ are the direction cosines of the given line L.
