10. Given two fixed points \( \mathrm{P}(-3,4) \) and \( \mathrm{Q}(5,-2) \); Find the coordinates of points A and B in PQ ; such 5 \( \mathrm{PA}=3 \mathrm{PQ} \) and \( 3 \mathrm{~PB}=2 \mathrm{PQ} \).

To find the coordinates of points \( A \) and \( B \) on the line segment \( PQ \) such that \( PA = 3PQ \) and \( 3PB = 2PQ \), we first need to determine the distance \( PQ \). 1. **Calculate the distance \( PQ \)**: The distance between points \( P(-3, 4) \) and \( Q(5, -2) \) can be calculated using the distance formula: \[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] where \( P(x_1, y_1) = (-3, 4) \) and \( Q(x_2, y_2) = (5, -2) \). \[ PQ = \sqrt{(5 - (-3))^2 + (-2 - 4)^2} = \sqrt{(5 + 3)^2 + (-2 - 4)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \] 2. **Find the coordinates of point \( A \)**: We know that \( PA = 3PQ \). Since \( PQ = 10 \), we have: \[ PA = 3 \times 10 = 30 \] This means point \( A \) is located 30 units away from point \( P \) in the direction of point \( Q \). To find the coordinates of point \( A \), we first find the unit vector in the direction from \( P \) to \( Q \): \[ \text{Direction vector } \vec{PQ} = (5 - (-3), -2 - 4) = (8, -6) \] The magnitude of this vector is \( PQ = 10 \), so the unit vector \( \hat{u} \) is: \[ \hat{u} = \left( \frac{8}{10}, \frac{-6}{10} \right) = \left( 0.8, -0.6 \right) \] Now, we can find point \( A \): \[ A = P + 30 \hat{u} = (-3, 4) + 30(0.8, -0.6) = (-3 + 24, 4 - 18) = (21, -14) \] 3. **Find the coordinates of point \( B \)**: We know that \( 3PB = 2PQ \). Since \( PQ = 10 \), we have: \[ PB = \frac{2}{3} \times 10 = \frac{20}{3} \] This means point \( B \) is located \( \frac{20}{3} \) units away from point \( P \) in the direction of point \( Q \). Using the same unit vector \( \hat{u} \): \[ B = P + \frac{20}{3} \hat{u} = (-3, 4) + \frac{20}{3}(0.8, -0.6) \] Calculating the components: \[ B = \left(-3 + \frac{20 \times 0.8}{3}, 4 - \frac{20 \times 0.6}{3}\right) = \left(-3 + \frac{16}{3}, 4 - \frac{12}{3}\right) \] Converting \(-3\) and \(4\) to fractions: \[ B = \left(-\frac{9}{3} + \frac{16}{3}, \frac{12}{3} - \frac{12}{3}\right) = \left(\frac{7}{3}, 0\right) \] 4. **Final coordinates**: The coordinates of points \( A \) and \( B \) are: \[ A(21, -14) \quad \text{and} \quad B\left(\frac{7}{3}, 0\right) \]