Question

Find the coordinates of point Q that lies along the directed line segment from R(-2, 4) to S(18, -6) and partitions the segment in the ratio of 3:7.
Please help!!

Answer 1

Check the picture below

thus then

$\bf \qquad \textit{internal division of a line segment}\\\\ R(-2,4)\qquad S(18,-6)\qquad ratio1=3\qquad ratio2=7\qquad 3:7\\ \quad \\ \quad \\ \cfrac{RQ}{QS}=\cfrac{ratio1}{ratio2}\implies \cfrac{R}{S}=\cfrac{3}{7} \implies 7R=3S \\\\\\ 7(-2,4)=3(18,-6)$

$\bf {{ Q=\left(\cfrac{\textit{sum of "x" values}}{ratio1+ratio2}\quad ,\quad \cfrac{\textit{sum of "y" values}}{ratio1+ratio2}\right)}}\\ \quad \\ \qquad thus\qquad \\ \quad \\ Q=\left(\cfrac{(7\cdot -2)+(3\cdot 18)}{3+7}\quad ,\quad \cfrac{(7\cdot 4)+(3\cdot -6)}{3+7}\right)$