The plane `2x+y+2z=9` intersects the coordinate axes at A,B,C.The orthocentre of the triangle ABC is

Options

A. `(2,1,2)`
B. `(3/2,3,3/2)`
C. `(3/13,1/13,3/13)`
D. `(9/7,27/7,9/7)`

Answer

To find the orthocenter of the triangle formed by the intersection of the plane \(2x + y + 2z = 9\) with the coordinate axes, we will follow these steps:

### Step 1: Find the points of intersection with the coordinate axes
To find the points where the plane intersects the coordinate axes, we set two variables to zero at a time and solve for the third variable.

1. **Intersection with the x-axis**: Set \(y = 0\) and \(z = 0\):
\[ 2x + 0 + 0 = 9 \implies 2x = 9 \implies x = \frac{9}{2} \]
So, the point \(A\) is \(\left(\frac{9}{2}, 0, 0\right)\).

2. **Intersection with the y-axis**: Set \(x = 0\) and \(z = 0\):
\[ 0 + y + 0 = 9 \implies y = 9 \]
So, the point \(B\) is \((0, 9, 0)\).

3. **Intersection with the z-axis**: Set \(x = 0\) and \(y = 0\):
\[ 0 + 0 + 2z = 9 \implies 2z = 9 \implies z = \frac{9}{2} \]
So, the point \(C\) is \((0, 0, \frac{9}{2})\).

### Step 2: Determine the coordinates of points A, B, and C
The coordinates of the points of intersection are:
- \(A\left(\frac{9}{2}, 0, 0\right)\)
- \(B(0, 9, 0)\)
- \(C(0, 0, \frac{9}{2})\)

### Step 3: Find the orthocenter of triangle ABC
The orthocenter of a triangle in 3D can be found using the formula for the orthocenter based on the vertices of the triangle. The orthocenter \(H\) can be calculated as follows:

Let \(P\), \(Q\), and \(R\) be the coordinates of the orthocenter. The coordinates can be derived from the condition that the sum of the perpendicular distances from the orthocenter to the sides of the triangle equals zero.

From the properties of the orthocenter:
- Let \(P = x\), \(Q = y\), \(R = z\).
- The relationship from the triangle vertices gives:
\[ \frac{P}{1} = \frac{Q}{\frac{1}{2}} = \frac{R}{1} \]
This implies:
\[ P = k, \quad Q = \frac{k}{2}, \quad R = k \]
for some constant \(k\).

### Step 4: Substitute into the plane equation
Substituting \(P\), \(Q\), and \(R\) into the plane equation:
\[ 2P + Q + 2R = 9 \]
Substituting the values of \(P\), \(Q\), and \(R\):
\[ 2k + \frac{k}{2} + 2k = 9 \]
Combining like terms:
\[ 4k + \frac{k}{2} = 9 \implies \frac{8k + k}{2} = 9 \implies \frac{9k}{2} = 9 \]
Multiplying both sides by 2:
\[ 9k = 18 \implies k = 2 \]

### Step 5: Find the coordinates of the orthocenter
Now substituting \(k = 2\) back into the expressions for \(P\), \(Q\), and \(R\):
\[ P = 2, \quad Q = \frac{2}{2} = 1, \quad R = 2 \]
Thus, the orthocenter \(H\) has coordinates:
\[ H(2, 1, 2) \]

### Final Answer
The orthocenter of triangle \(ABC\) is at the point \((2, 1, 2)\).

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