Step-by-step explanation:
A,B, and C are collinear, and B is between A and C. The ratio of AB to BC is 1:1 of A IS AT (-1,9) and B (2,0)
to find out point C use section formula
[tex](\frac{mx_2+nx_1}{m+n} ,\frac{my_2+ny_1}{m+n} )[/tex]
A is (-1,9) that is our (x1,y1)
that is our (x2,y2)
ratio is 1:1 that is m and n
Plug in the values in the formula
[tex](\frac{mx_2+nx_1}{m+n} ,\frac{my_2+ny_1}{m+n} )[/tex]
[tex](\frac{1(x_2)+1(-1)}{1+1} ,\frac{1(y_2)+1(9)}{1+1} ) =(2,0)\\\frac{1(x_2)+1(-1)}{1+1}=2\\\frac{1(x_2)+1(-1)}{2}=2\\\\x_2-1=4\\x_2= 5\\\frac{1(y_2)+1(9)}{1+1}=0 \\\frac{1(y_2)+1(9)}{2} =0\\\\y_2+9=0\\x_2= -9[/tex]
Answer C is (5,-9)
