Two vertices of a triangle are `(4,-3) & (-2, 5)`. If the orthocentre of the triangle is at `(1,2)`, find coordinates of the third vertex .

Options

A. `(-33,-26)`
B. `(33,26)`
C. `(26,33)`
D. none of these

Explanation

To find the coordinates of the third vertex of the triangle given two vertices and the orthocenter, we can follow these steps:

### Step 1: Define the vertices and orthocenter
Let the two vertices of the triangle be:
- Vertex A: \( A(4, -3) \)
- Vertex B: \( B(-2, 5) \)
- Let the third vertex be \( C(h, k) \)
- The orthocenter \( H \) is given as \( H(1, 2) \)

### Step 2: Find the slope of line AB
The slope \( m_{AB} \) of line segment AB can be calculated using the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the coordinates of points A and B:
\[ m_{AB} = \frac{5 - (-3)}{-2 - 4} = \frac{5 + 3}{-6} = \frac{8}{-6} = -\frac{4}{3} \]

### Step 3: Find the slope of line AH
The slope \( m_{AH} \) of line segment AH (from A to H) is:
\[ m_{AH} = \frac{2 - (-3)}{1 - 4} = \frac{2 + 3}{1 - 4} = \frac{5}{-3} = -\frac{5}{3} \]

### Step 4: Use the property of perpendicular lines
Since the altitude from A to BC is perpendicular to line BC, we can use the relationship that the product of the slopes of two perpendicular lines is -1. Thus:
\[ m_{AB} \cdot m_{BC} = -1 \]
Let the slope of line BC be \( m_{BC} \). Therefore:
\[ -\frac{4}{3} \cdot m_{BC} = -1 \implies m_{BC} = \frac{3}{4} \]

### Step 5: Find the slope of line BC
The slope of line BC can also be expressed in terms of the coordinates of C:
\[ m_{BC} = \frac{k - 5}{h + 2} \]
Setting this equal to the previously found slope:
\[ \frac{k - 5}{h + 2} = \frac{3}{4} \]
Cross-multiplying gives:
\[ 4(k - 5) = 3(h + 2) \implies 4k - 20 = 3h + 6 \implies 3h - 4k + 26 = 0 \quad \text{(Equation 1)} \]

### Step 6: Find the slope of line AH
Using the slope of line AH:
\[ m_{AH} = \frac{k - (-3)}{h - 4} = \frac{k + 3}{h - 4} \]
Setting this equal to the slope of line AB:
\[ \frac{k + 3}{h - 4} = -\frac{5}{3} \]
Cross-multiplying gives:
\[ 3(k + 3) = -5(h - 4) \implies 3k + 9 = -5h + 20 \implies 5h + 3k - 11 = 0 \quad \text{(Equation 2)} \]

### Step 7: Solve the system of equations
Now we have two equations:
1. \( 3h - 4k + 26 = 0 \)
2. \( 5h + 3k - 11 = 0 \)

We can solve these equations simultaneously. From Equation 1, we can express \( h \) in terms of \( k \):
\[ 3h = 4k - 26 \implies h = \frac{4k - 26}{3} \]

Substituting this into Equation 2:
\[ 5\left(\frac{4k - 26}{3}\right) + 3k - 11 = 0 \]
Multiplying through by 3 to eliminate the fraction:
\[ 5(4k - 26) + 9k - 33 = 0 \implies 20k - 130 + 9k - 33 = 0 \implies 29k - 163 = 0 \implies k = \frac{163}{29} = 5.62 \]

Substituting \( k \) back to find \( h \):
\[ h = \frac{4(5.62) - 26}{3} = \frac{22.48 - 26}{3} = \frac{-3.52}{3} = -1.17 \]

### Final Result
Thus, the coordinates of the third vertex \( C \) are \( (h, k) \approx (-1.17, 5.62) \).