By using determinants , find the value of 'y' for which the points `(1,3),(2,5)` and `(3,y)` are collinear.

Answer

To find the value of 'y' for which the points (1,3), (2,5), and (3,y) are collinear, we can use the concept of determinants. The points are collinear if the area of the triangle formed by them is zero.

### Step-by-Step Solution:

1. **Set Up the Determinant**:
The area of the triangle formed by the points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the determinant:
\[ \text{Area} = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \]
For our points \((1, 3)\), \((2, 5)\), and \((3, y)\), we can substitute:
\[ \text{Area} = \frac{1}{2} \begin{vmatrix} 1 & 3 & 1 \\ 2 & 5 & 1 \\ 3 & y & 1 \end{vmatrix} \]

2. **Calculate the Determinant**:
We need to calculate the determinant:
\[ \begin{vmatrix} 1 & 3 & 1 \\ 2 & 5 & 1 \\ 3 & y & 1 \end{vmatrix} \]
Using the formula for the determinant of a 3x3 matrix:
\[ = 1 \cdot (5 \cdot 1 - y \cdot 1) - 3 \cdot (2 \cdot 1 - 3 \cdot 1) + 1 \cdot (2y - 15) \]
Simplifying this gives:
\[ = 1(5 - y) - 3(2 - 3) + (2y - 15) \]
\[ = 5 - y + 3 + 2y - 15 \]
\[ = (5 + 3 - 15) + (-y + 2y) \]
\[ = -7 + y \]

3. **Set the Area to Zero**:
For the points to be collinear, the area must be zero:
\[ \frac{1}{2}(-7 + y) = 0 \]
This implies:
\[ -7 + y = 0 \]
Therefore:
\[ y = 7 \]

### Conclusion:
The value of \(y\) for which the points (1,3), (2,5), and (3,y) are collinear is \(y = 7\).