Question

Point C (4, 2) divides the line segment joining points A (2, -1) and B (x, y) such that AC: CB = 3.1 what are the coordinates of point B?

Answer 1

The coordinates of B are $(\frac{14}{3} ,3)$.

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 $P(x,y)$ divides the line segment $A(x_1,y_1)$ and $B(x_2,y_2)$ in the ratio

m : n are $\left(\frac{\mathbf{m} \mathbf{x}_{2}+\mathbf{n} \mathbf{x}_{\mathbf{1}}}{\mathbf{m}+\mathbf{n}}, \frac{\mathbf{m} \mathbf{y}_{2}+\mathbf{n} \mathbf{y}_{\mathbf{1}}}{\mathbf{m}+\mathbf{n}}\right)$

Here, $x_1=2, \ x_2=x, \ x_3=4, \ y_1=-1, \ y_2=y, \ y_3=2$ and m = 3, n = 1

Using section formula,

$C(x_3,y_3)=\left(\frac{mx_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)$

Substitute the given values in the section formula.

$C(4,2)=\left(\frac{3 \times x+1 \times 2}{3+1}, \frac{3 \times y+1 \times (-1)}{3+1}\right)$

$C(4,2)=\left(\frac{3x+2}{4}, \frac{3y-1}{4}\right)$

Equate the x-coordinates and y-coordinates.

$\frac{3x+2}{4}=4, \ \ \ \frac{3y-1}{4}=2$

$\ 3x+2=16, \ \ \ \ 3y-1=8$

$3x=14, \ \ \ \ 3y=9$

$x=\frac{14}{3}, \ \ \ \ y=3$

Hence the coordinates of B are $(\frac{14}{3} ,3)$.