explain what a directed line segment is and describe how you would find the coordinates of Point P along the directed line segment a b that partitions so that the ratio of AP to PB is 1:3​

Answer:Please read the explanation as the answer says to explain/describe so it isn't a short answer.Given the line segment AB; divide it into 1+3 parts since we want the ratio of AP to PB to be 1 to 3.Start at A in count over until you have landed on the next marker after A. You have found P since the rest is 3 parts of the 4.I'm not too sure if we are looking for algebra or something like this above in bold.Step-by-step explanation:A directed line segment is a segment with direction and distance.But let's assume point A is at [tex](x_1,y_1)[/tex] and B is at [tex](x_3,y_3)[/tex].We are looking for point P at [tex](x_3,y_3)[/tex] such that the ratio of AP to PB is 1 to 3.So there are 4 parts where 1 part of is the length from A to P and the other 3 parts are the length of PB. Now those aren't the unit measurement of the line segments. The length of AP is 1k units and the length of PB is 3k units long.Anyways, to find the point [tex](x_3,y_3)[/tex], I'm going to graph my points and form 2 triangles (one inside a bigger one).  These triangles will be similar since they have all three angles corresponding angles are congruent.I tried to draw my graphs so that AP made up 1 part and PB made up the other 3 parts of my 4 parted line segment.  I parted into 4 equal lengths by eyeballing it using those little green markers.  I want P to be located on the first green marker reading from left to right since AP is 1 part of that 4 parter there.So in similar triangles corresponding sides are proportional:[tex]\frac{x_3-x_1}{x_2-x_1}=\frac{y_3-y_1}{y_2-y_1}=\frac{1k}{4k}[/tex]Remember we are solving for P in terms of the other numbers here that are not [tex]x_3 \text{ and } y_3[/tex].I'm going to solve this one for [tex]x_3[/tex]:[tex]\frac{x_3-x_1}{x_2-x_1}=\frac{1k}{4k}[/tex]The k's on the right hand fraction cancel:[tex]\frac{x_3-x_1}{x_2-x_1}=\frac{1}{4}[/tex]Cross multiply:[tex]4(x_3-x_1)=1(x_2-x_1)[/tex]Distribute:[tex]4x_3-4x_1=x_2-x_1[/tex]Add [tex]4x_1[/tex] on both side:[tex]4x_3=x_2+3x_1[/tex]Divide both sides by 4:[tex]x_3=\frac{x_2+3x_1}{4}[/tex]Let's solve the other equation for [tex]y_3[/tex]:[tex]\frac{y_3-y_1}{y_2-y_1}=\frac{1k}{4k}[/tex]The k's on the right hand fraction cancel:[tex]\frac{y_3-y_1}{y_2-y_1}=\frac{1}{4}[/tex]Cross multiply:[tex]4(y_3-y_1)=1(y_2-y_1)[/tex]Distribute:[tex]4y_3-4y_1=y_2-y_1[/tex]Add [tex]4y_1[/tex]:[tex]4y_3=y_2+3y_1[/tex]Divide both sides by 4:[tex]y_3=\frac{y_2+3y_1}{4}[/tex]So point P is at [tex](\frac{x_2+3x_1}{4},\frac{y_2+3y_1}{4})[/tex].Let's verify that AP to PB is 1 to 3 by using the distance formula:We are going to compute the distance from A to P:[tex](x_1,y_1) \text{ to } (\frac{x_2+3x_1}{4},\frac{y_2+3y_1}{4})[/tex].[tex]\sqrt{(\frac{x_2+3x_1}{4}-x_1)^2+(\frac{y_2+3y_1}{4}-y_1)^2}[/tex]Recall 3/4 -  1  = -1/4:[tex]\sqrt{(\frac{x_2}{4}-\frac{x_1}{4})^2+(\frac{y_2}{4}-\frac{y_1}{4})^2[/tex][tex]\sqrt{(\frac{x_2-x_1}{4})^2+(\frac{y_2-y_1}{4})^2[/tex]Factoring out [tex]\sqrt{\frac{1}{4^2}}[/tex]:[tex]\sqrt{\frac{1}{4^2}}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex][tex]\frac{1}{4}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]So this says right here that AP's distance is [tex]\frac{1}{4}[/tex] the distance of AB which is what was intended.Let's see if we get from PB 3/4 the length of AB.So now we are computing the distance from P to B:[tex](\frac{x_2+3x_1}{4},\frac{y_2+3y_1}{4}) \text{ to } (x_2,y_2)[/tex][tex]\sqrt{(x_2-\frac{x_2+3x_1}{4})^2+(y_2-\frac{y_2+3y_1}{4})^2}[/tex]Recall 1  -  1/4 =3/4:[tex]\sqrt{(\frac{3}{4}x_2-\frac{3}{4}x_1)^2+(\frac{3}{4}y_2-\frac{3}{4}y_1)^2[/tex]Factoring out [tex]\sqrt{(\frac{3}{4})^2}[/tex]:[tex]\sqrt{(\frac{3}{4})^2}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex][tex]\frac{3}{4}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]So again this says PB is 3/4 the length of AB.Just so we are clear the distance from A to B:[tex](x_1,y_1)[/tex] to [tex](x_2,y_2)[/tex] is given by the formula:[tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex].----------Below here contains a solution for one dimension.So again divide the line segments into 4 parts.You want AP to be one part of this 4 parts.You want PB to be three parts of this 4 parts.The distance from A to P is [tex]x_3-x_1[/tex] or [tex]1k[/tex].The distance from P to B is [tex]x_2-x_3[/tex] or [tex]3k[/tex].The distance from A to B is[tex]x_2-x_1[/tex] is [tex]4k[/tex].We want AP to PB to be the ratio 1 to 3.So we want:[tex]\frac{x_3-x_1}{x_2-x_1}=\frac{1k}{4k}[/tex].Simplify right hand side:[tex]\frac{x_3-x_1}{x_2-x_1}=\frac{1}{4}[/tex]Cross multiply:[tex]4(x_3-x_1)=1(x_2-x_1)[/tex]Distribute:[tex]4x_3-4x_1=x_2-x_1[/tex]Add [tex]4x_1[/tex] on both sides:[tex]4x_3=x_2+3x_1[/tex]Divide both sides by 4:[tex]x_3=\frac{x_2+3x_1}{4}[/tex].Given the line segment AB; divide it into 1+3 parts since we want the ratio of AP to PB to be 1 to 3.Start at A in count over until you have landed on the next marker after A. You have found P since the rest is 3 parts of the 4.