\[ \text { Diagram } 3 \] Koordinat bagi lebah \( P \) dan lebah \( Q \) masing-masing adalah \( (-4,4) \) dan \( (8,-12) \). Kedua- dua lebah itu terbang ke arah satu sama lain pada satu garis lurus dengan halaju berbeza. Halaju lebah \( P \) adalah tiga kali ganda halaju lebah \( Q \). Cari jarak lebah \( P \) dari titik asalnya apabila ia bertemu dengan lebah \( Q \). The coordinates of bee \( P \) and bee \( Q \) are \( (-4,4) \) and \( (8,-12) \) respectively. Both bees fly towards each other on a straight line with different velocity. The velocity of bee \( P \) is three times more than velocity of bee \( Q \). Find the distance of bee \( P \) from its initial point when it meets with bee \( Q \). \( [3 \) markah / marks]

To solve the problem, we first need to determine the distance between the two bees \( P \) and \( Q \) and then find the point where they meet. 1. **Calculate the distance between \( P \) and \( Q \)**: The coordinates of bee \( P \) are \( (-4, 4) \) and the coordinates of bee \( Q \) are \( (8, -12) \). The distance \( d \) between the two points can be calculated using the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] where \( (x_1, y_1) = (-4, 4) \) and \( (x_2, y_2) = (8, -12) \). Substituting the values: \[ d = \sqrt{(8 - (-4))^2 + (-12 - 4)^2} \] \[ = \sqrt{(8 + 4)^2 + (-12 - 4)^2} \] \[ = \sqrt{(12)^2 + (-16)^2} \] \[ = \sqrt{144 + 256} \] \[ = \sqrt{400} \] \[ = 20 \] 2. **Determine the velocities of the bees**: Let the velocity of bee \( Q \) be \( v \). Then the velocity of bee \( P \) is \( 3v \). 3. **Calculate the time taken for the bees to meet**: Since they are moving towards each other, the relative velocity is: \[ v_{\text{relative}} = v + 3v = 4v \] The time \( t \) taken for them to meet can be calculated using the formula: \[ t = \frac{\text{Distance}}{\text{Relative Velocity}} = \frac{20}{4v} = \frac{5}{v} \] 4. **Calculate the distance traveled by bee \( P \)**: The distance \( d_P \) traveled by bee \( P \) is given by: \[ d_P = \text{velocity of } P \times t = 3v \times t = 3v \times \frac{5}{v} = 15 \] Thus, the distance of bee \( P \) from its initial point when it meets with bee \( Q \) is \( \boxed{15} \).