3. What are the coordinates of the point \( \frac{1}{4} \) of the way from \( A(-6,-3) \) to \( B(10,9) \) ?(

To find the coordinates of the point that is \(\frac{1}{4}\) of the way from point \( A(-6, -3) \) to point \( B(10, 9) \), you can use the **section formula** for a point dividing a line segment internally in a given ratio. ### Step-by-Step Solution: 1. **Understand the Section Formula:** If you have two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \), the coordinates of a point \( P \) that divides the line segment \( AB \) in the ratio \( m:n \) is given by: \[ P\left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right) \] In this problem, since the point \( P \) is \(\frac{1}{4}\) of the way from \( A \) to \( B \), the ratio \( m:n \) is \( 1:3 \). This is because \( \frac{1}{4} \) of the distance from \( A \) means that \( P \) is closer to \( A \) by a ratio of \( 1 \) part towards \( B \) and \( 3 \) parts remaining. 2. **Apply the Section Formula:** Given: \[ A(-6, -3) \quad \text{and} \quad B(10, 9) \] Ratio \( m:n = 1:3 \) Plugging into the section formula: \[ P_x = \frac{1 \times 10 + 3 \times (-6)}{1 + 3} = \frac{10 - 18}{4} = \frac{-8}{4} = -2 \] \[ P_y = \frac{1 \times 9 + 3 \times (-3)}{1 + 3} = \frac{9 - 9}{4} = \frac{0}{4} = 0 \] 3. **Conclusion:** The coordinates of the point \( P \) that is \(\frac{1}{4}\) of the way from \( A(-6, -3) \) to \( B(10, 9) \) are: \[ P(-2, 0) \] ### Alternative Method: Using Vector Addition You can also approach this problem using vector addition. The idea is to move \(\frac{1}{4}\) of the displacement from \( A \) to \( B \). 1. **Find the Displacement Vector from \( A \) to \( B \):** \[ \vec{AB} = (10 - (-6), 9 - (-3)) = (16, 12) \] 2. **Find \(\frac{1}{4}\) of the Displacement:** \[ \frac{1}{4} \vec{AB} = \left( \frac{16}{4}, \frac{12}{4} \right) = (4, 3) \] 3. **Add This to Point \( A \) to Find \( P \):** \[ P = A + \frac{1}{4} \vec{AB} = (-6 + 4, -3 + 3) = (-2, 0) \] Both methods yield the same result. ### Final Answer: The coordinates of the point are \( \boxed{\,(-2,\;\,0)\,} \).