Find the co-ordinates of the incentre of the triangle formed by the lines `y-15=0`, `12y-5x=0` and `4y+3x=0`.

Answer

To find the coordinates of the incenter of the triangle formed by the lines \(y - 15 = 0\), \(12y - 5x = 0\), and \(4y + 3x = 0\), we will follow these steps:

### Step 1: Identify the equations of the lines
The equations of the lines are:
1. Line 1: \(y = 15\)
2. Line 2: \(12y - 5x = 0\) or \(y = \frac{5}{12}x\)
3. Line 3: \(4y + 3x = 0\) or \(y = -\frac{3}{4}x\)

### Step 2: Find the points of intersection (vertices of the triangle)
#### Point A (intersection of Line 1 and Line 3):
Set \(y = 15\) in \(4y + 3x = 0\):
\[ 4(15) + 3x = 0 \implies 60 + 3x = 0 \implies 3x = -60 \implies x = -20 \]
So, Point A is \((-20, 15)\).

#### Point B (intersection of Line 1 and Line 2):
Set \(y = 15\) in \(12y - 5x = 0\):
\[ 12(15) - 5x = 0 \implies 180 - 5x = 0 \implies 5x = 180 \implies x = 36 \]
So, Point B is \((36, 15)\).

#### Point C (intersection of Line 2 and Line 3):
Set \(12y - 5x = 0\) and \(4y + 3x = 0\):
From \(4y + 3x = 0\), we have \(y = -\frac{3}{4}x\).
Substituting into \(12y - 5x = 0\):
\[ 12\left(-\frac{3}{4}x\right) - 5x = 0 \implies -9x - 5x = 0 \implies -14x = 0 \implies x = 0 \]
Then substituting \(x = 0\) into \(y = -\frac{3}{4}x\):
\[ y = 0 \]
So, Point C is \((0, 0)\).

### Step 3: Calculate the lengths of the sides of the triangle
Let \(a\), \(b\), and \(c\) be the lengths of the sides opposite to vertices A, B, and C respectively.
- Length \(a\) (BC):
\[ a = \sqrt{(36 - 0)^2 + (15 - 0)^2} = \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39 \]

- Length \(b\) (AC):
\[ b = \sqrt{(-20 - 0)^2 + (15 - 0)^2} = \sqrt{(-20)^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25 \]

- Length \(c\) (AB):
\[ c = \sqrt{(36 - (-20))^2 + (15 - 15)^2} = \sqrt{(36 + 20)^2 + 0^2} = \sqrt{56^2} = 56 \]

### Step 4: Calculate the coordinates of the incenter
The coordinates of the incenter \((h, k)\) are given by:
\[ h = \frac{a x_1 + b x_2 + c x_3}{a + b + c}, \quad k = \frac{a y_1 + b y_2 + c y_3}{a + b + c} \]
Substituting the coordinates of points A \((-20, 15)\), B \((36, 15)\), and C \((0, 0)\):
\[ h = \frac{39(-20) + 25(36) + 56(0)}{39 + 25 + 56} = \frac{-780 + 900 + 0}{120} = \frac{120}{120} = 1 \]
\[ k = \frac{39(15) + 25(15) + 56(0)}{39 + 25 + 56} = \frac{585 + 375 + 0}{120} = \frac{960}{120} = 8 \]

### Final Answer
The coordinates of the incenter of the triangle are \((1, 8)\).

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