Question

in what ratio does the point (-4,6) divide the line segment joining the points A(-6,10) and B(3,-8)?

Answer 1

Answer:

x = mx2 + nx1/m+n

-4= k3 + (-1)/k+1b. (let the ratio be k:1)

-4k -4 = 3k -1

-7k = 3

k = 3/-1

or 3:1 externally

Answer 2

☯ Let the points A(-6,10) and B(3,-8) is divided by point (-4,6) in the ratio m : n.

DIAGRAM

$\setlength{\unitlength}{14mm}\begin{picture}(7,5)(0,0)\thicklines\put(0,0){\line(1,0){5}}\put(5.1, - 0.3){\sf B}\put( - 0.2, - 0.3){\sf A}\put(5.2, 0){\sf (3,-8)}\put( - 0.7, 0){\sf (-6,10)}\put(2.3, 0.2){\sf C}\put(2.2, - 0.3){\sf (-4,6)}\put(5, 0){\circle*{0.1}}\put(2.4, 0){\circle*{0.1}}\put(0, 0){\circle*{0.1}}\put(1,0.2){\sf m}\put(3.5, 0.2){\sf n}\end{picture}$

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$\begin{gathered}\underline{\bigstar\:\boldsymbol{Using\:section\:formula\::}}\\ \\\end{gathered}$

$\begin{gathered}\star\;{\boxed{\sf{\pink{(x,y) = \bigg( \dfrac{m x_2 + n x_1}{m + n}\;,\; \dfrac{m y_2 + n y_1}{m + n} \bigg)}}}}\\ \\\end{gathered}$

$\begin{gathered}\sf Here \begin{cases} \sf{x_1 , y_1 = -6,10} \\ \sf{x_2 , y_2 = 3,-8} \end{cases}\\ \\\end{gathered}$

Therefore,⠀⠀

$\begin{gathered}:\implies\sf \dfrac{m \times 3 + n \times -6}{m + n} = -4 \\\\\\:\implies\sf m \times 3 + n \times -6 = - 4m -4n\\\\\\:\implies\sf 3m - 6n = -4m - 4n\\\\\\:\implies\sf 3m + 4m = 6n - 4n \\\\\\:\implies\sf 7m = 2n\\\\\\:\implies\sf \dfrac{m}{n} = \dfrac{ 2}{7}\\\\\\:\implies{\underline{\boxed{\sf{\purple{m : n = 2 : 7}}}}}\;\bigstar\\ \\\end{gathered}$

$\therefore\;{\underline{\sf{The\;ratio\; in \;which \;(-4,6)\; divides\; the \;line\; segment\; is\; {\textsf{\textbf{2 : 7}}}.}}}$

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