Given :
A graph .
To Find :
A point C that divides the segment into the ratio AC:AB= 3/4 .
Solution :
From the graph we can see that point A and B are ( 1 , -2 ) and ( 5 , 6 ) respectively .
Now , we know point which divide a line segment with end point [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] in m : n is :
[tex]x=\dfrac{mx_2+nx_1}{m+n}\ ,\ y=\dfrac{my_2+ny_1}{m+n}[/tex]
Putting all given values we get :
[tex]x=\dfrac{3(5)+4(1)}{3+4}\ ,\ y=\dfrac{3(6)+4(-2)}{3+4}\\\\x=\dfrac{19}{7}\ ,\ y =\dfrac{10}{7}[/tex]
Therefore , point C is [tex](\dfrac{19}{7},\dfrac{10}{7})[/tex] .
Hence , this is the required solution .
