Question

given the points A(-3,-5) and B (5,0), find the coordinates of the point P on a directed line segment AB that partitions AB in the ratio 2:3

Answer 1

$\bf ~~~~~~~~~~~~\textit{internal division of a line segment} \\\\\\ A(-3,-5)\qquad B(5,0)\qquad \qquad \stackrel{\textit{ratio from A to B}}{2:3} \\\\\\ \cfrac{A\underline{P}}{\underline{P} B} = \cfrac{2}{3}\implies \cfrac{A}{B} = \cfrac{2}{3}\implies 3A=2B\implies 3(-3,-5)=2(5,0)\\\\[-0.35em] \rule{31em}{0.25pt}\\\\ P=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em] \rule{31em}{0.25pt}$

$\bf P=\left(\cfrac{(3\cdot -3)+(2\cdot 5)}{2+3}\quad ,\quad \cfrac{(3\cdot -5)+(2\cdot 0)}{2+3}\right) \\\\\\ P=\left( \cfrac{-9+10}{5}~~,~~\cfrac{-15+0}{5} \right)\implies P=\left(\frac{1}{5}~,~-3 \right)$

Answer 2

Answer:

$(\frac{1}{5},-3)$.

Step-by-step explanation:

We have been given the coordinates of points A(-3,-5) and B (5,0). We are asked to find the coordinates of the point P on a directed line segment AB that partitions AB in the ratio 2:3.  

We will use section formula to solve our given problem. When a point P divides a line segment internally in ratio m:n, then coordinates of point P are:

$[x=\frac{m\cdot x_2+n\cdot x_1}{m+n};y=\frac{m\cdot y_2+n\cdot y_1}{m+n}]$

Upon substituting coordinates of our given points in section formula we will get,

$[x=\frac{2\cdot 5+3\cdot -3}{2+3};y=\frac{2\cdot 0+3\cdot -5}{2+3}]$

$[x=\frac{10-9}{5};y=\frac{0-15}{5}]$

$[x=\frac{1}{5};y=\frac{-15}{5}]$

$[x=\frac{1}{5};y=-3]$

Therefore, the coordinates of point P are $(\frac{1}{5},-3)$.