Find the centre of the inscribed circle of the triangle the equations, of whose sides are `y-15=0, 12y+5x=0 and 4y-3x=0`.

Answer

To find the center of the inscribed circle (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 points of intersection of the lines
We need to find the vertices of the triangle by determining the points of intersection of the given lines.

1. **Intersection of \(y - 15 = 0\) and \(12y + 5x = 0\)**:
- From \(y - 15 = 0\), we have \(y = 15\).
- Substitute \(y = 15\) into \(12y + 5x = 0\):
\[ 12(15) + 5x = 0 \implies 180 + 5x = 0 \implies 5x = -180 \implies x = -36 \]
- So, the point \(B\) is \((-36, 15)\).

2. **Intersection of \(y - 15 = 0\) and \(4y - 3x = 0\)**:
- Again, from \(y - 15 = 0\), we have \(y = 15\).
- Substitute \(y = 15\) into \(4y - 3x = 0\):
\[ 4(15) - 3x = 0 \implies 60 - 3x = 0 \implies 3x = 60 \implies x = 20 \]
- So, the point \(C\) is \((20, 15)\).

3. **Intersection of \(12y + 5x = 0\) and \(4y - 3x = 0\)**:
- From \(12y + 5x = 0\), we can express \(x\) in terms of \(y\):
\[ x = -\frac{12}{5}y \]
- Substitute this into \(4y - 3x = 0\):
\[ 4y - 3\left(-\frac{12}{5}y\right) = 0 \implies 4y + \frac{36}{5}y = 0 \]
\[ \left(4 + \frac{36}{5}\right)y = 0 \implies \frac{20 + 36}{5}y = 0 \implies \frac{56}{5}y = 0 \implies y = 0 \]
- Substitute \(y = 0\) back into \(x = -\frac{12}{5}y\):
\[ x = 0 \]
- So, the point \(A\) is \((0, 0)\).

### Step 2: Calculate the lengths of the sides of the triangle
Let the vertices be:
- \(A(0, 0)\)
- \(B(-36, 15)\)
- \(C(20, 15)\)

Now, we calculate the lengths of the sides:
- Length \(BC\) (denoted as \(a\)):
\[ a = \sqrt{((-36) - 20)^2 + (15 - 15)^2} = \sqrt{(-56)^2 + 0^2} = 56 \]

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

- Length \(AB\) (denoted as \(c\)):
\[ c = \sqrt{((-36) - 0)^2 + (15 - 0)^2} = \sqrt{(-36)^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39 \]

### Step 3: Use the incenter formula
The coordinates of the incenter \((x_{ic}, y_{ic})\) are given by:
\[ x_{ic} = \frac{a x_1 + b x_2 + c x_3}{a + b + c} \]
\[ y_{ic} = \frac{a y_1 + b y_2 + c y_3}{a + b + c} \]

Substituting the values:
- \(x_1 = 0\), \(y_1 = 0\)
- \(x_2 = -36\), \(y_2 = 15\)
- \(x_3 = 20\), \(y_3 = 15\)

Calculating \(x_{ic}\):
\[ x_{ic} = \frac{56(0) + 25(-36) + 39(20)}{56 + 25 + 39} = \frac{0 - 900 + 780}{120} = \frac{-120}{120} = -1 \]

Calculating \(y_{ic}\):
\[ y_{ic} = \frac{56(0) + 25(15) + 39(15)}{56 + 25 + 39} = \frac{0 + 375 + 585}{120} = \frac{960}{120} = 8 \]

### Final Result
Thus, the coordinates of the incenter are:
\[ \text{Incenter} = (-1, 8) \]