Question

Plot a point C that divides the segment into the ratio AC:AB= 3/4

Answer 1

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 $(x_1,y_1)$ and $(x_2,y_2)$ in m : n is :

$x=\dfrac{mx_2+nx_1}{m+n}\ ,\ y=\dfrac{my_2+ny_1}{m+n}$

Putting all given values we get :

$x=\dfrac{3(5)+4(1)}{3+4}\ ,\ y=\dfrac{3(6)+4(-2)}{3+4}\\\\x=\dfrac{19}{7}\ ,\ y =\dfrac{10}{7}$

Therefore , point C is $(\dfrac{19}{7},\dfrac{10}{7})$ .

Hence , this is the required solution .