MATH SOLVE

4 months ago

Q:
# Given a line segment with endpoints A(16, 8) and B(1, 3) what are the coordinates of the line segment partitioned two-fifths from A to B?

Accepted Solution

A:

check the picture below.

since it's being cut in 2/5 thus a ratio of 2:5, which means that AB is being cut into 2+5 or 7 even pieces, AP takes 2 of those and PB takes 5 of those.

Also notice that "from A to B" matters greatly, since that means AP takes the smaller ratio, whilst the other takes the other, let's check then,

[tex]\bf \left. \qquad \right.\textit{internal division of a line segment} \\\\\\ A(16,8)\qquad B(1,3)\qquad \qquad 2:5 \\\\\\ \cfrac{AP}{PB} = \cfrac{2}{5}\implies \cfrac{A}{B} = \cfrac{2}{5}\implies 5A=2B\implies 5(16,8)=2(1,3) \\\\ -------------------------------\\\\[/tex]

[tex]\bf { P=\left(\cfrac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \cfrac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)}\\\\ -------------------------------\\\\ P=\left(\cfrac{(5\cdot 16)+(2\cdot 1)}{2+5}\quad ,\quad \cfrac{(5\cdot 8)+(2\cdot 3)}{2+5}\right) \\\\\\ P=\left( \cfrac{80+2}{7}~~,~~\cfrac{40+6}{7} \right)\implies P=\left(\cfrac{82}{7}~~,~~\cfrac{46}{7} \right) \\\\\\ P=\left(11\frac{5}{7}~~,~~6\frac{4}{7} \right)[/tex]

since it's being cut in 2/5 thus a ratio of 2:5, which means that AB is being cut into 2+5 or 7 even pieces, AP takes 2 of those and PB takes 5 of those.

Also notice that "from A to B" matters greatly, since that means AP takes the smaller ratio, whilst the other takes the other, let's check then,

[tex]\bf \left. \qquad \right.\textit{internal division of a line segment} \\\\\\ A(16,8)\qquad B(1,3)\qquad \qquad 2:5 \\\\\\ \cfrac{AP}{PB} = \cfrac{2}{5}\implies \cfrac{A}{B} = \cfrac{2}{5}\implies 5A=2B\implies 5(16,8)=2(1,3) \\\\ -------------------------------\\\\[/tex]

[tex]\bf { P=\left(\cfrac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \cfrac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)}\\\\ -------------------------------\\\\ P=\left(\cfrac{(5\cdot 16)+(2\cdot 1)}{2+5}\quad ,\quad \cfrac{(5\cdot 8)+(2\cdot 3)}{2+5}\right) \\\\\\ P=\left( \cfrac{80+2}{7}~~,~~\cfrac{40+6}{7} \right)\implies P=\left(\cfrac{82}{7}~~,~~\cfrac{46}{7} \right) \\\\\\ P=\left(11\frac{5}{7}~~,~~6\frac{4}{7} \right)[/tex]