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What are the coordinates of point [tex]\( P \)[/tex] on the line segment from [tex]\( A \)[/tex] to [tex]\( B \)[/tex] such that [tex]\( P \)[/tex] is [tex]\(\frac{1}{4}\)[/tex] the length of the line segment from [tex]\( A \)[/tex] to [tex]\( B \)[/tex]?

A. [tex]\(\left(\frac{-29}{4}, \frac{-3}{2}\right)\)[/tex]
B. [tex]\(\left(\frac{-13}{4}, \frac{1}{2}\right)\)[/tex]
C. [tex]\(\left(\frac{-11}{4}, \frac{-1}{2}\right)\)[/tex]
D. [tex]\(\left(\frac{25}{4}, \frac{-1}{2}\right)\)[/tex]

Sagot :

GinnyAnswer
To determine the coordinates of point [tex]\( P \)[/tex] such that it lies [tex]\(\frac{1}{4}\)[/tex] of the way along the line segment from [tex]\( A \)[/tex] to [tex]\( B \)[/tex], we need the coordinates of points [tex]\( A \)[/tex] and [tex]\( B \)[/tex]. Let's use [tex]\( A \left(\frac{-29}{4}, \frac{-3}{2}\right) \)[/tex] and [tex]\( B \left(\frac{25}{4}, \frac{-1}{2}\right) \)[/tex].

### Step 1: Calculate the distance vector [tex]\(\overrightarrow{AB}\)[/tex]
First, we find the vector [tex]\(\overrightarrow{AB}\)[/tex]. This vector is obtained by subtracting the coordinates of [tex]\( A \)[/tex] from [tex]\( B \)[/tex]:

[tex]\[ \overrightarrow{AB} = \left( \frac{25}{4} - \frac{-29}{4}, \frac{-1}{2} - \frac{-3}{2} \right) \][/tex]

Simplify each component:

[tex]\[ \overrightarrow{AB}_x = \frac{25}{4} + \frac{29}{4} = \frac{54}{4} = 13.5 \][/tex]
[tex]\[ \overrightarrow{AB}_y = \frac{-1}{2} + \frac{3}{2} = 1.0 \][/tex]

So, the vector [tex]\(\overrightarrow{AB}\)[/tex] is:

[tex]\[ \overrightarrow{AB} = (13.5, 1.0) \][/tex]

### Step 2: Calculate the coordinates of point [tex]\( P \)[/tex]
Point [tex]\( P \)[/tex] is located [tex]\(\frac{1}{4}\)[/tex] of the way from [tex]\( A \)[/tex] to [tex]\( B \)[/tex]. To find the coordinates of [tex]\( P \)[/tex], we start at point [tex]\( A \)[/tex] and move [tex]\(\frac{1}{4}\)[/tex] of the way along the vector [tex]\(\overrightarrow{AB}\)[/tex]:

[tex]\[ P_x = A_x + \frac{1}{4} \cdot \overrightarrow{AB}_x \][/tex]
[tex]\[ P_y = A_y + \frac{1}{4} \cdot \overrightarrow{AB}_y \][/tex]

Using the values we have:

[tex]\[ P_x = \frac{-29}{4} + \frac{1}{4} \cdot 13.5 \][/tex]
[tex]\[ P_y = \frac{-3}{2} + \frac{1}{4} \cdot 1.0 \][/tex]

Simplify these expressions:

[tex]\[ P_x = \frac{-29}{4} + \frac{13.5}{4} = \frac{-29 + 13.5}{4} = \frac{-15.5}{4} = -3.875 \][/tex]

[tex]\[ P_y = \frac{-3}{2} + \frac{1}{4} \cdot 1 = \frac{-3}{2} + \frac{1}{4} = \frac{-6}{4} + \frac{1}{4} = \frac{-5}{4} = -1.25 \][/tex]

Therefore, the coordinates of point [tex]\( P \)[/tex] are:

[tex]\[ P(-3.875, -1.25) \][/tex]
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