If two vertices of a triangle are `(-2,3)` and `(5,-1)` the orthocentre lies at the origin, and the centroid on the line `x+y=7` , then the third vertex lies at

Options

A. (7,4)
B. (8,14)
C. (12,21)
D. none of these

Explanation

To solve the problem, we need to find the coordinates of the third vertex of a triangle given the following conditions:

1. Two vertices of the triangle are A(-2, 3) and B(5, -1).
2. The orthocenter (H) lies at the origin (0, 0).
3. The centroid (G) lies on the line x + y = 7.

Let the coordinates of the third vertex C be (h, k).

### Step 1: Find the Centroid

The formula for the centroid \( G \) of a triangle with vertices \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) is given by:

\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]

Substituting the known coordinates:

\[ G = \left( \frac{-2 + 5 + h}{3}, \frac{3 - 1 + k}{3} \right) = \left( \frac{3 + h}{3}, \frac{2 + k}{3} \right) \]

### Step 2: Set Up the Equation for the Centroid

Since the centroid lies on the line \( x + y = 7 \), we can substitute the coordinates of the centroid into this equation:

\[ \frac{3 + h}{3} + \frac{2 + k}{3} = 7 \]

Multiplying through by 3 to eliminate the denominator:

\[ 3 + h + 2 + k = 21 \]

This simplifies to:

\[ h + k = 16 \quad \text{(Equation 1)} \]

### Step 3: Find the Slopes for the Orthocenter

The slopes of the lines from the vertices to the orthocenter must be perpendicular.

1. The slope of line \( AH \) from \( A(-2, 3) \) to \( H(0, 0) \):

\[ \text{slope of } AH = \frac{0 - 3}{0 + 2} = \frac{-3}{2} \]

2. The slope of line \( BH \) from \( B(5, -1) \) to \( H(0, 0) \):

\[ \text{slope of } BH = \frac{0 + 1}{0 - 5} = \frac{1}{-5} = -\frac{1}{5} \]

3. The slope of line \( BC \) from \( B(5, -1) \) to \( C(h, k) \):

\[ \text{slope of } BC = \frac{k + 1}{h - 5} \]

### Step 4: Use the Perpendicularity Condition

Since \( AH \) and \( BC \) are perpendicular, we have:

\[ \frac{-3}{2} \cdot \frac{k + 1}{h - 5} = -1 \]

This simplifies to:

\[ \frac{3(k + 1)}{2(h - 5)} = 1 \]

Cross-multiplying gives:

\[ 3(k + 1) = 2(h - 5) \]

Expanding this:

\[ 3k + 3 = 2h - 10 \]

Rearranging gives:

\[ 2h - 3k = 13 \quad \text{(Equation 2)} \]

### Step 5: Solve the System of Equations

Now we have two equations:

1. \( h + k = 16 \) (Equation 1)
2. \( 2h - 3k = 13 \) (Equation 2)

From Equation 1, we can express \( k \) in terms of \( h \):

\[ k = 16 - h \]

Substituting this into Equation 2:

\[ 2h - 3(16 - h) = 13 \]

Expanding and simplifying:

\[ 2h - 48 + 3h = 13 \]

Combining like terms:

\[ 5h - 48 = 13 \]

Adding 48 to both sides:

\[ 5h = 61 \]

Dividing by 5:

\[ h = \frac{61}{5} \]

### Step 6: Find \( k \)

Substituting \( h \) back into Equation 1:

\[ k = 16 - \frac{61}{5} = \frac{80}{5} - \frac{61}{5} = \frac{19}{5} \]

### Final Result

Thus, the coordinates of the third vertex \( C \) are:

\[ C\left(\frac{61}{5}, \frac{19}{5}\right) \]