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

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UpStudy AI · Accepted Answer

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 $$ \boxed{\,(-2,\;\,0)\,} $$.
The point is at $$(-2, 0)$$.

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