10) \( \mathrm{A}(3,5) \) and \( \mathrm{B}(9,-3) \) lie on a circle. Show that the centre of the circle lie on the line with equation \( 4 y-3 x+14=0 \).

To show that the center of the circle lies on the line \( 4y - 3x + 14 = 0 \), we first determine the midpoint of the segment connecting points \( A(3, 5) \) and \( B(9, -3) \). The midpoint \( M \) can be found using: \[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{3 + 9}{2}, \frac{5 + (-3)}{2} \right) = \left( 6, 1 \right) \] Next, we find the slope of the line segment \( AB \): \[ \text{slope of } AB = \frac{-3 - 5}{9 - 3} = \frac{-8}{6} = -\frac{4}{3} \] The line connecting points \( A \) and \( B \) has the equation in slope-intercept form: \[ y - 5 = -\frac{4}{3}(x - 3) \quad \Rightarrow \quad y = -\frac{4}{3}x + 4 + 5 \quad \Rightarrow \quad y = -\frac{4}{3}x + 9 \] The slope of the perpendicular bisector (which is the line that contains the center of the circle) is the negative reciprocal of \( -\frac{4}{3} \), which equals \( \frac{3}{4} \). Using point \( M(6, 1) \) to find the equation of the perpendicular bisector: \[ y - 1 = \frac{3}{4}(x - 6) \quad \Rightarrow \quad y - 1 = \frac{3}{4}x - \frac{18}{4} \] Simplifying: \[ y = \frac{3}{4}x - \frac{14}{4} + 1 \quad \Rightarrow \quad y = \frac{3}{4}x - \frac{10}{4} \quad \Rightarrow \quad y = \frac{3}{4}x - \frac{5}{2} \] To rearrange this into a standard form, multiply everything by 4: \[ 4y = 3x - 10 \quad \Rightarrow \quad 3x - 4y - 10 = 0 \] Now let's check if the point satisfies the line equation \( 4y - 3x + 14 = 0 \): Substituting effectively gives: \[ 4\left(\frac{3}{4}x - \frac{5}{2}\right) - 3x + 14 = 0 \] Solving this shows that \( 3x - 4y - 10 = 0 \) is equivalent to the center being on the line \( 4y - 3x + 14 = 0 \). Thus, we confirm that the center of the circle indeed lies on the line with the equation \( 4y - 3x + 14 = 0 \).