Line segment PQ is a directed line segment beginning at P(6, -5) and ending at Q(-2, 4).

Find point R on the line segment PQ that partitions it into the segments PR and RQ in the ratio 3:2.

A. [tex]\(\left(\frac{6}{5}, \frac{2}{5}\right)\)[/tex]

B. [tex]\(\left(\frac{14}{5}, -\frac{7}{5}\right)\)[/tex]

C. [tex]\(\left(-\frac{6}{5}, \frac{2}{5}\right)\)[/tex]

D. [tex]\(\left(\frac{14}{5}, \frac{7}{5}\right)\)[/tex]



Answer :

To find the coordinates of point [tex]\( R \)[/tex] that divides the line segment [tex]\( PQ \)[/tex] in the ratio [tex]\( 3:2 \)[/tex], we use the section formula. The section formula states that if a point [tex]\( R \)[/tex] divides the line segment joining two points [tex]\( P(x_1, y_1) \)[/tex] and [tex]\( Q(x_2, y_2) \)[/tex] in the ratio [tex]\( m:n \)[/tex], then the coordinates of [tex]\( R \)[/tex] are given by:

[tex]\[ R\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right) \][/tex]

Given:
- [tex]\( P = (6, -5) \)[/tex]
- [tex]\( Q = (-2, 4) \)[/tex]
- Ratio [tex]\( m:n = 3:2 \)[/tex]

We can now apply the section formula to find [tex]\( R \)[/tex].

### Calculating the x-coordinate of [tex]\( R \)[/tex]:
[tex]\[ R_x = \frac{m x_2 + n x_1}{m + n} \][/tex]

Substituting the values:
[tex]\[ R_x = \frac{3(-2) + 2(6)}{3 + 2} \][/tex]

Simplifying:
[tex]\[ R_x = \frac{3(-2) + 2(6)}{5} \\ R_x = \frac{-6 + 12}{5} \\ R_x = \frac{6}{5} \][/tex]

### Calculating the y-coordinate of [tex]\( R \)[/tex]:
[tex]\[ R_y = \frac{m y_2 + n y_1}{m + n} \][/tex]

Substituting the values:
[tex]\[ R_y = \frac{3(4) + 2(-5)}{3 + 2} \][/tex]

Simplifying:
[tex]\[ R_y = \frac{3(4) + 2(-5)}{5} \\ R_y = \frac{12 - 10}{5} \\ R_y = \frac{2}{5} \][/tex]

Hence, the coordinates of point [tex]\( R \)[/tex] are:

[tex]\[ R \left(\frac{6}{5}, \frac{2}{5}\right) \][/tex]

From the given choices, the correct option is:

A. [tex]\(\left(\frac{6}{5}, \frac{2}{5}\right)\)[/tex]