Let the incentre of `DeltaABC` is I(2, 5). If `A=(1, 13)` and `B=(-4, 1)`, then the coordinates of C are

Options

A. `(1, 10)`
B. `(10, 1)`
C. `(8, 2)`
D. `(9, 2)`

Answer

To find the coordinates of point C in triangle ABC, given the incentre I(2, 5), point A(1, 13), and point B(-4, 1), we can follow these steps:

### Step 1: Find the equation of line AB
We can use the coordinates of points A and B to find the equation of line AB.

1. The slope \( m \) of line AB is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 13}{-4 - 1} = \frac{-12}{-5} = \frac{12}{5} \]

2. Using point-slope form \( y - y_1 = m(x - x_1) \):
\[ y - 1 = \frac{12}{5}(x + 4) \]
Simplifying this, we get:
\[ y - 1 = \frac{12}{5}x + \frac{48}{5} \]
\[ y = \frac{12}{5}x + \frac{53}{5} \]
Multiplying through by 5 to eliminate the fraction:
\[ 5y = 12x + 53 \implies 12x - 5y + 53 = 0 \]

### Step 2: Calculate the distance from the incentre to line AB
The distance \( d \) from point I(2, 5) to line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Where \( A = 12, B = -5, C = 53 \) and \( (x_0, y_0) = (2, 5) \).

Calculating:
\[ d = \frac{|12(2) - 5(5) + 53|}{\sqrt{12^2 + (-5)^2}} = \frac{|24 - 25 + 53|}{\sqrt{144 + 25}} = \frac{|52|}{13} = 4 \]
This distance is equal to the radius \( R \) of the incircle.

### Step 3: Find the slopes of lines AC and BC
Let the coordinates of C be \( (h, k) \).

1. The slope of line BC:
\[ m_{BC} = \frac{k - 1}{h + 4} \]

2. The slope of line AC:
\[ m_{AC} = \frac{k - 13}{h - 1} \]

### Step 4: Use the distance from the incentre to line BC
Using the same distance formula for line BC:
\[ d = \frac{|12h - 5k + 53|}{\sqrt{12^2 + (-5)^2}} = 4 \]
This gives us:
\[ |12h - 5k + 53| = 4 \cdot 13 = 52 \]
This leads to two equations:
1. \( 12h - 5k + 53 = 52 \)
2. \( 12h - 5k + 53 = -52 \)

### Step 5: Solve the equations
From the first equation:
\[ 12h - 5k = -1 \quad \text{(1)} \]

From the second equation:
\[ 12h - 5k = -105 \quad \text{(2)} \]

### Step 6: Use the slopes to find k
Using the slope of BC:
\[ \frac{k - 1}{h + 4} = 0 \implies k = 1 \]

### Step 7: Substitute k into equation (1)
Substituting \( k = 1 \) into equation (1):
\[ 12h - 5(1) = -1 \implies 12h - 5 = -1 \implies 12h = 4 \implies h = \frac{1}{3} \]

### Step 8: Final coordinates
Thus, the coordinates of point C are:
\[ C\left(\frac{1}{3}, 1\right) \]

### Summary of Steps
1. Find the equation of line AB.
2. Calculate the distance from the incentre to line AB.
3. Find the slopes of lines AC and BC.
4. Use the distance from the incentre to line BC.
5. Solve the equations obtained.
6. Substitute to find the coordinates of C.