Question

Find the coordinates of point S that lies along the directed line segment from R(-14, -1) to T(4, -13) and partitions the segment in the ratio of 1:5.

Answer 1

Check the picture below.

I'd like to point out, is 1:5 from R to T, since that matters, that way we know the ratio from RS is the 1 and the ratio ST is the 5.

$\bf \left. \qquad \right.\textit{internal division of a line segment} \\\\\\ R(-14,-1)\qquad T(4,-13)\qquad \qquad 1:5 \\\\\\ \cfrac{RS}{ST} = \cfrac{1}{5}\implies \cfrac{R}{T} = \cfrac{1}{5}\implies 5R=1T \implies 5(-14,-1)=1(4,-13)\\\\ -------------------------------$

$\bf { S=\left(\cfrac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \cfrac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)}\\\\ -------------------------------\\\\ S=\left(\cfrac{(5\cdot -14)+(1\cdot 4)}{1+5}\quad ,\quad \cfrac{(5\cdot -1)+(1\cdot -13)}{1+5}\right)$

Answer 2

Answer:  The required co-ordinates of the point S are (-11, -6).

Step-by-step explanation:  We are given to find the co-ordinates of the point S that lies along the directed line segment from R(-14, -1) to T(4, -13) and partitions the segment in the ratio of 1:5.

We know that

the co-ordinates of a point that divides the line segment joining the points (a, b) and (c, d) in the ratio m : n is given by

$\left(\dfrac{mc+na}{m+n},\dfrac{md+nb}{m+n}\right).$

According to the given information, we have

(a, b) = (-14, -1),  (c, d) = (4, -13)  and  m : n = 1 : 5.

Therefore, the co-ordinates of the point S are given by

$\left(\dfrac{1\times4+5\times(-14)}{1+5},\dfrac{1\times(-13)+5\times(-1)}{}\right)\\\\\\=\left(\dfrac{4-70}{6},\dfrac{-13-5}{6}\right)\\\\\\=\left(\dfrac{-66}{6},\dfrac{-18}{6}\right)\\\\=(-11,-3)$

Thus, the required co-ordinates of the point S are (-11, -6).