Question

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

Answer 1

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Q1) Find the ratio in which line segment joining the points (-3,10) and (6,-8) is divided by (-1,6)

$\huge{\mathtt{\purple{A}\green{N}\pink{S}\blue{W}\purple{E}\green{R}\pink{!}\blue{!}}}$

$\: \: \twoheadrightarrow \boxed { \bf \pink{ \: to \: find \: the \: ratio}}$

$\gg\sf{Consider\: the \: point (-3,10) \: is \:point A \:and \: B \: as\: a (6,8)point}$

$\dagger \boxed { \bf{ \: by \: using \: section \: formula}}$

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

$\star \sf{substitute \: the \: given \: values}$

$\: \: \implies \bf{ - 1 = \frac{m( - 3) + n(6)}{m + n} } \: \: \: \: \: and \: \: \: 6 = \frac{m(10) + n(8)}{m + n}$

$\: \: \implies \bf{ - 1 = \frac{m( - 3) + n( 6)}{m + n} }$

$\: \: \implies \bf{ - m - n = - 3m + 6n}$

$\: \: \implies \bf{ - m - n + 3m - 6n = 0}$

$\: \: \implies \bf{2m - 7n = 0}.......((1))$

$\pink \ast \to \bf{6 = \frac{m(10) + n( - 8)}{m + n} }$

$\: \: \mapsto \bf{ - 6m - 6n = 10m - 8n}$

$\: \mapsto \bf{ - 6m - 6n - 10m + 8n = 0}$

$\: \mapsto \bf{ - 16m + 2n = 0}.......(2)$

$\: \: \boxed { \rm \red{ \: multiply \: in \: eqn(1) \: by \: 8 \: we \: get \: 16m - 14n = 0}}$

$\: \: \to \bf{16m -2n = 0}....(2)$

$\: \to \bf{16m - 14n = 0}.......(3)$

$\: \: \: \bigstar \underline{\bf{ \: subtracting \: both \: equations}}$

$\: \implies \bf{0 + 12m = 0}$

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

$\implies \bf16( 0) - 14n = 0$

$\: \implies \bf{n = 0}$

Answer 2

HIII MATE.... ur answer is attached...