# Find the ratio in which the line segment joining the points (?3,10) and (6,?8) is divided by (?1,6).

$Using\;the\;section\;formula,\;if\;a\;point\;(x,y)\;divides\;the\;line\;joining\; th e\\*points\;(x_1,y_1)\;and\;(x_2,y_2)\;in\;the\;ratio\;m:n,\;then\\*(x,y)=(\frac {mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})\\*Let\;the\;point\;P(-1,6)\; divides\;the\;line\;joining\;A(-3,10)\;and\;(6,-8)\;in\;the\\*ratio\;k:1,\;then ,\; by\;section\;formula,\\*\Rightarrow (-1,6)=(\frac {6k-3}{k+1},\frac{-8k+10}{k+1})\\*\therefore \frac{6k-3}{k+1}=-1\\*\Rightarrow 6k-3=-k-1\\* \Rightarrow 6k+k=3-1\\*\Rightarrow 7k=2\\*\therefore k=\frac{2}{7}\\* Hence,\;the\;point\;P\;divides\;AB\;in\;the\;ratio\;2:7$

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