Question 1115405
 Find the point dividing the interval AB in the ratio {{{1:3}}} if A({{{-1}}},{{{1}}}) and B is ({{{3}}},{{{-1}}}) 

To find the point {{{P}}} that divides a segment {{{AB}}} into a particular ratio, determine the ratio {{{k}}} by writing the numerator over the sum of the numerator and the denominator of the given ratio. 
Next, find the rise and the run (slope) of the line. Finally, add {{{k}}} times the run to the {{{x}}}-coordinate of {{{A}}} and add {{{k}}} times the rise to the {{{y}}}-coordinate of {{{A}}}. 

This process is summarized with the following formula:


P({{{x}}},{{{y}}})=({{{x[1]+k(x[2]-x[1])}}} , {{{y[1]+k(y[2]-y[1])}}} )

you have:

{{{k=1/3}}}

A({{{-1}}},{{{1}}}) => {{{x[1]=-1}}}, {{{y[1]=1}}}
and 
B is ({{{3}}},{{{-1}}}) => {{{x[2]= 3}}}, {{{y[2]=-1}}}


({{{x}}},{{{y}}})=({{{-1+(1/3)(3-(-1))}}} , {{{1+(1/3)(-1-1)}}} )

({{{x}}},{{{y}}})=({{{-1+(1/3)(3+1)}}} , {{{1+(1/3)(-2)}}} )

({{{x}}},{{{y}}})=({{{-1+4/3}}} , {{{1-2/3}}} )

({{{x}}},{{{y}}})=({{{-3/3+4/3}}} , {{{3/3-2/3}}} )

({{{x}}},{{{y}}})=({{{1/3}}} ,{{{ 1/3}}} )


{{{drawing( 600, 600, -5, 5, -5, 5,
line(-1,1,3,-1),
circle(-1,1,.05), locate(-1,1,A(-1,1)),
circle(3,-1,.05), locate(3,-1,B(3,-1)),
circle(1/3,1/3,.05), locate(1/3,1/3,P(1/3,1/3)),
 graph( 600, 600, -5, 5, -5, 5, 0)) }}}