MATH SOLVE

8 months ago

Q:
# Number 8 please, What are the coordinates of point C below segment AC that is partitioned at point B in a ratio of 2 to 5.

Accepted Solution

A:

[tex]\bf \left. \qquad \right.\textit{internal division of a line segment}
\\\\\\
A(5,0)\qquad C(x,y)\qquad
\qquad 2:5
\\\\\\
\cfrac{AB}{BC} = \cfrac{2}{5}\implies \cfrac{A}{C}=\cfrac{2}{5}\implies 5A=2C\implies 5(5,0)=2(x,y)\\\\
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{ B=\left(\cfrac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \cfrac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)}[/tex]

[tex]\bf -------------------------------\\\\ B=\left(\cfrac{(5\cdot 5)+(2\cdot x)}{2+5}\quad ,\quad \cfrac{(5\cdot 0)+(2\cdot y)}{2+5}\right)=\boxed{(8,0)} \\\\\\ B=\left( \cfrac{25+2x}{7}~~,~~\cfrac{0+2y}{7} \right)=(8,0)\implies \begin{cases} \cfrac{25+2x}{7}=8\\\\ 25+2x=56\\ 2x=31\\\\ \boxed{x=\cfrac{31}{2}}\\ -------\\ \cfrac{0+2y}{7}=0\\\\ 2y=0\\ \boxed{y=0} \end{cases}[/tex]

[tex]\bf -------------------------------\\\\ B=\left(\cfrac{(5\cdot 5)+(2\cdot x)}{2+5}\quad ,\quad \cfrac{(5\cdot 0)+(2\cdot y)}{2+5}\right)=\boxed{(8,0)} \\\\\\ B=\left( \cfrac{25+2x}{7}~~,~~\cfrac{0+2y}{7} \right)=(8,0)\implies \begin{cases} \cfrac{25+2x}{7}=8\\\\ 25+2x=56\\ 2x=31\\\\ \boxed{x=\cfrac{31}{2}}\\ -------\\ \cfrac{0+2y}{7}=0\\\\ 2y=0\\ \boxed{y=0} \end{cases}[/tex]