Question

find the ratio in which (11,15) divides the line segment joining the points (15,5) and (9,10)​

Answer 1

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✯Find the ratio in which (11,15) divides the line segment joining the points (15,5) and

(9,10).

$\underline{ \bf \color{black}{ \red \bigstar \: find \: the \: ratio }}$

$\: \large \underline{ \sf{ \color{lime} { \: \dag \: by \: \: section \: \: formula}}}$

$\implies\bf{x=\dfrac{mx_1+nx_2}{m+n} \: and \: y=\dfrac{my_1+ny_2}{m+n}}$

$\mapsto\large\sf{X=\dfrac{mx_1+mx_2}{m+n}}$

$\to\large\sf{11=\dfrac{m(15)+n(9)}{m+n}</p><p></p><p>[tex]\to\large\sf{11m+11n=15m+9n}}$

$\to\large\sf{11m-15m+11n-9n=0}$

$\to\large\sf{-4m+2n=0}$

$\to\large\sf{-4m=-2n}$

$\to\large\sf{2m=n}$

$\star \color{orange} \sf{2m - n = 0....(1)}$

$\mapsto\large\sf{Y=\dfrac{my_1+ny_2}{m+n}}$

$\to\large\sf{15=\dfrac{5m+10n}{m+n}}$

$\to\large\sf{15m+15n=5m+10n}$

$\to\large\sf{15m+15n-5m-10n=0}$

$\to\large\sf{10m+5n=0}$

$to\large\sf{10m=-5n}$

$\to\large\sf{-2m=n}$

$\: \star \color{orange} \sf{2m + n = 0}......(2)$

$\bigstar \boxed{ \underline{ \underline { \bf{subtracting \: equation \: (1) \: and \: (2)}}}}$

$\: \bf{2m - n = 0}$

$\ \bf{2m + n = 0}$

$\: \boxed{ \rm{n = 0}}$

$\: \bf{ substitute \: in \: equation \: (1)}$

$\: \bf{2m - 0 = 0}$

$\: \bf{2m = 0}$

$\: \bf{ie \: m = 0}$