To find the coordinates of point \( X \) on segment \( AB \) such that the ratio of \( AX \) to \( XB \) is \( 3:5 \), we can use the section formula.
Let the coordinates of point \( A \) be \( A(-1, 2) \) and the coordinates of point \( B \) be \( B(3, 6) \). The ratio of \( AX \) to \( XB \) is given as \( 3:5 \), meaning \( AX = 3k \) and \( XB = 5k \) for some \( k \). Therefore, the total length \( AB \) is \( AX + XB = 3k + 5k = 8k \).
Using the section formula, the coordinates of point \( X \) which divides the line segment \( AB \) in the ratio \( m:n \) is given by:
\[ X = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \]
Here \( m = 3 \), \( n = 5 \), \( A(x_1, y_1) = (-1, 2) \), and \( B(x_2, y_2) = (3, 6) \).
Substituting these values:
\[ X_x = \frac{3(3) + 5(-1)}{3+5} = \frac{9 - 5}{8} = \frac{4}{8} = \frac{1}{2} \]
\[ X_y = \frac{3(6) + 5(2)}{3+5} = \frac{18 + 10}{8} = \frac{28}{8} = \frac{7}{2} \]
Thus, the coordinates of point \( X \) are:
\[ X\left(\frac{1}{2}, \frac{7}{2}\right) \]
So, the coordinates of point \( X \) on line segment \( AB \) such that the ratio of \( AX \) to \( XB \) is \( 3:5 \) are \( \left( \frac{1}{2}, \frac{7}{2} \right) \).