Given:
Line of the graph.
AB : BC = 3 : 4
To find:
The point B on AC.
Solution:
The coordinate of A is (8, -5).
The coordinate of C is (1, 9).
Using section formula:
[tex]$B(x, y)=\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)[/tex]
Here, [tex]x_1=8, y_1=-5, x_2=1, y_2=9[/tex] and m = 3, n = 4
[tex]$B(x, y)=\left(\frac{3\times 1 +4 \times 8}{3+4}, \frac{3 \times 9+4\times (-5)}{3+4}\right)[/tex]
[tex]$B(x, y)=\left(\frac{3 +32}{7}, \frac{27-20}{7}\right)[/tex]
[tex]$B(x, y)=\left(\frac{35}{7}, \frac{7}{7}\right)[/tex]
[tex]$B(x, y)=\left(5, 1)[/tex]
The point B on AC is (5, 1).
