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:
[tex]B = (\frac{14}{3},3)[/tex]
Step-by-step explanation:
Given
[tex]C = (4,2)[/tex]
[tex]A = (2,-1)[/tex]
[tex]AC : CB = 3 : 1[/tex]
Required
Find the coordinates of B
Coordinates of a line segment is calculated using:
[tex](x,y) = (\frac{mx_2+nx_1}{n+m}, \frac{my_2 + ny_1}{n+m})[/tex]
In this case:
[tex](x,y) = (4,2)[/tex]
[tex]AC:CB = m : n = 3:1[/tex]
[tex]A = (2,-1)[/tex] --- [tex](x_1,y_1)[/tex]
The equation becomes
[tex](4,2) = (\frac{3x_2+2}{1+3}, \frac{3y_2 - 1}{1+3})[/tex]
So, the coordinates of B is: [tex](x_2,y_2)[/tex]
Solving further
[tex](4,2) = (\frac{3x_2+2}{4}, \frac{3y_2 - 1}{4})[/tex]
Multiply through by 4
[tex]4 * (4,2) = (\frac{3x_2+2}{4}, \frac{3y_2 - 1}{4}) * 4[/tex]
[tex](16,8) = (3x_2+2, 3y_2 - 1)[/tex]
By comparison:
[tex]3x_2 + 2 = 16[/tex]
[tex]3y_2 - 1 = 8[/tex]
So:
[tex]3x_2 + 2 = 16[/tex]
[tex]3x_2 = 16 - 2[/tex]
[tex]3x_2 = 14[/tex]
[tex]x_2 = \frac{14}{3}[/tex]
[tex]3y_2 - 1 = 8[/tex]
[tex]3y_2 = 8+1[/tex]
[tex]3y_2 = 9[/tex]
[tex]y_2 = 3[/tex]
The coordinates of B is:
[tex]B = (\frac{14}{3},3)[/tex]
