To find the point [tex]\( R \)[/tex] that partitions the directed line segment [tex]\( PQ \)[/tex] in the ratio [tex]\( PR:RQ = 3:2 \)[/tex], we use the section formula. According to the section formula, if we have a line segment with endpoints [tex]\( P(x_1, y_1) \)[/tex] and [tex]\( Q(x_2, y_2) \)[/tex], and we want to find the point [tex]\( R(x, y) \)[/tex] that divides the line segment in the ratio [tex]\( m:n \)[/tex], the coordinates of [tex]\( R \)[/tex] are given by:
[tex]\[
R \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right)
\][/tex]
For [tex]\( P(6, -5) \)[/tex] and [tex]\( Q(-2, 4) \)[/tex], and the ratio [tex]\( 3:2 \)[/tex] (i.e., [tex]\( m = 3 \)[/tex] and [tex]\( n = 2 \)[/tex]):
1. Calculate the x-coordinate of [tex]\( R \)[/tex]:
[tex]\[
x_R = \frac{3 \cdot (-2) + 2 \cdot 6}{3 + 2}
\][/tex]
[tex]\[
x_R = \frac{ -6 + 12 }{ 5 }
\][/tex]
[tex]\[
x_R = \frac{ 6 }{ 5 }
\][/tex]
2. Calculate the y-coordinate of [tex]\( R \)[/tex]:
[tex]\[
y_R = \frac{3 \cdot 4 + 2 \cdot (-5)}{3 + 2}
\][/tex]
[tex]\[
y_R = \frac{ 12 - 10 }{ 5 }
\][/tex]
[tex]\[
y_R = \frac{ 2 }{ 5 }
\][/tex]
Therefore, the coordinates of point [tex]\( R \)[/tex] that partitions the segment [tex]\( PQ \)[/tex] in the given ratio are:
[tex]\[ R \left( \frac{6}{5}, \frac{2}{5} \right) \][/tex]
So, the correct answer is:
D. [tex]\( \left( \frac{6}{5}, \frac{2}{5} \right) \)[/tex]
