The coordinates of B are [tex](\frac{14}{3} ,3)[/tex].
Solution:
Given A(2, –1), B(x, y) and C(4, 2)
AB is a line segment C is a point on AB.
AC : CB = 3 : 1
To find the coordinates of B:
Section formula:
The point [tex]P(x,y)[/tex] divides the line segment [tex]A(x_1,y_1)[/tex] and [tex]B(x_2,y_2)[/tex] in the ratio
m : n are [tex]\left(\frac{\mathbf{m} \mathbf{x}_{2}+\mathbf{n} \mathbf{x}_{\mathbf{1}}}{\mathbf{m}+\mathbf{n}}, \frac{\mathbf{m} \mathbf{y}_{2}+\mathbf{n} \mathbf{y}_{\mathbf{1}}}{\mathbf{m}+\mathbf{n}}\right)[/tex]
Here, [tex]x_1=2, \ x_2=x, \ x_3=4, \ y_1=-1, \ y_2=y, \ y_3=2[/tex] and m = 3, n = 1
Using section formula,
[tex]$C(x_3,y_3)=\left(\frac{mx_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)[/tex]
Substitute the given values in the section formula.
[tex]$C(4,2)=\left(\frac{3 \times x+1 \times 2}{3+1}, \frac{3 \times y+1 \times (-1)}{3+1}\right)[/tex]
[tex]$C(4,2)=\left(\frac{3x+2}{4}, \frac{3y-1}{4}\right)[/tex]
Equate the x-coordinates and y-coordinates.
[tex]$\frac{3x+2}{4}=4, \ \ \ \frac{3y-1}{4}=2[/tex]
[tex]$\ 3x+2=16, \ \ \ \ 3y-1=8[/tex]
[tex]$ 3x=14, \ \ \ \ 3y=9[/tex]
[tex]$ x=\frac{14}{3}, \ \ \ \ y=3[/tex]
Hence the coordinates of B are [tex](\frac{14}{3} ,3)[/tex].
