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

$\huge{\underline{\underline {\bf{\blue{Answer:-}}}}}$

$\large\bold{\underline{\sf{\pink{\dag \; GivEn:-}}}}$

• Coordinates of A → (-6,10)
• Coordinates of B → (3,-8)
• Point dividing the line segment → C(-4,6)

$\large\bold{\underline{\sf{\pink{\dag \; Solutions:-}}}}$

$\implies Using \ section \ formula:-$

$x = \dfrac{ \sf m_1 \sf x_2 + \sf m_2 \sf x_1 }{ \sf m_1 + \sf m_2 }$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ or$

$y = \dfrac{ \sf m_1 \sf x_1 + \sf m_2 \sf y_1 }{ \sf m_1 + \sf m_2 }$

$\implies -4 = \dfrac{ 3 \sf m_1 - 6 \sf m_2 }{ \sf m_1 + \sf m_2 }$

$\implies -4 \sf m_1 - 4 \sf m_2 = 3 \sf m_1 - 6 \sf m_2$

$\implies -4 \sf m_1 - 3 \sf m_1 = -6 \sf m_2 + 4 m_2$

$\implies -7 \sf m_1 = -2 \sf m_2$

$\implies \dfrac{ \sf m_1}{ \sf m_2} = \dfrac{-7}{-2}$

$\implies \sf m_1 : \sf m_2 = 7 : 2$

★ To verify our answer also solve it from y point :-

$\implies 6 = \dfrac{ -8 \sf m_1 +10 \sf m_2 }{ \sf m_1 + \sf m_2 }$

$\implies 6 \sf m_1 + 6 \sf m_2 =-8 \sf m_1 +10 \sf m_2$

$\implies 6 \sf m_1 + 8 \sf m_1 = 10 \sf m_2 - 6 \sf m_2$

$\implies 14 \sf m_1 = 4 \sf m_2$

$\implies \dfrac{ \sf m_1}{ \sf m_2} = \cancel \dfrac{14}{4}$

$\implies \dfrac{ \sf m_1}{ \sf m_2} = \dfrac{7}{2}$

$\implies \sf m_1 : \sf m_2 = 7 : 2$

$\large\bold{\underline{\sf{\red{\dag \; Hence, \; in \; both \; case \; ratio \; is \; 7:2.}}}}$

$\dag \ Therefore, \ our \ ans \ is \ correct.$

$\large\bold{\underline{\boxed{\sf{\blue{ \sf m_1 : \sf m_2 \; = \; 7:2.}}}}}$

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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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