Question 3 (Essay Worth 10 points)

A point [tex]\( T \)[/tex] on a segment with endpoints [tex]\( D (1,4) \)[/tex] and [tex]\( F (7,1) \)[/tex] partitions the segment in a [tex]\( 3:1 \)[/tex] ratio. Find [tex]\( T \)[/tex]. You must show all work to receive credit.



Answer :

To solve the problem of finding the point [tex]\( T \)[/tex] that partitions the segment with endpoints [tex]\( D(1, 4) \)[/tex] and [tex]\( F(7, 1) \)[/tex] in a [tex]\( 3:1 \)[/tex] ratio, we will use the section formula for internal division of a line segment in a given ratio.

The section formula states that if a point [tex]\( T \)[/tex] divides the line segment joining two points [tex]\( D(x_1, y_1) \)[/tex] and [tex]\( F(x_2, y_2) \)[/tex] in the ratio [tex]\( m:n \)[/tex], then the coordinates of [tex]\( T \)[/tex] are given by:

[tex]\[ T_x = \frac{m \cdot x_2 + n \cdot x_1}{m+n} \][/tex]
[tex]\[ T_y = \frac{m \cdot y_2 + n \cdot y_1}{m+n} \][/tex]

Given:
- [tex]\( D(1, 4) \)[/tex] implies [tex]\( x_1 = 1 \)[/tex], [tex]\( y_1 = 4 \)[/tex]
- [tex]\( F(7, 1) \)[/tex] implies [tex]\( x_2 = 7 \)[/tex], [tex]\( y_2 = 1 \)[/tex]
- The ratio is [tex]\( m:n = 3:1 \)[/tex], which means [tex]\( m = 3 \)[/tex] and [tex]\( n = 1 \)[/tex]

First, we will find the x-coordinate of point [tex]\( T \)[/tex]:

[tex]\[ T_x = \frac{3 \cdot 7 + 1 \cdot 1}{3+1} \][/tex]
[tex]\[ T_x = \frac{21 + 1}{4} \][/tex]
[tex]\[ T_x = \frac{22}{4} \][/tex]
[tex]\[ T_x = 5.5 \][/tex]

Next, we calculate the y-coordinate of point [tex]\( T \)[/tex]:

[tex]\[ T_y = \frac{3 \cdot 1 + 1 \cdot 4}{3+1} \][/tex]
[tex]\[ T_y = \frac{3 + 4}{4} \][/tex]
[tex]\[ T_y = \frac{7}{4} \][/tex]
[tex]\[ T_y = 1.75 \][/tex]

Thus, the coordinates of point [tex]\( T \)[/tex] that partitions the segment joining [tex]\( D(1, 4) \)[/tex] and [tex]\( F(7, 1) \)[/tex] in a [tex]\( 3:1 \)[/tex] ratio are:

[tex]\[ T = (5.5, 1.75) \][/tex]