. In what ratio does the point (-4 ,6) divide the line segment joining the points (-6, 10) and (3, 6).​

1 thought on “. In what ratio does the point (-4 ,6) divide the line segment joining the points (-6, 10) and (3, 6).​”

1. [tex]\huge{\underline{\textsf{\textbf{\red{Question…}}}}}[/tex]

   In what ratio does the point [tex]\sf (-4,6)[/tex] divide the line segment joining the points [tex]\sf (-6,10)[/tex] and [tex]\sf (3,6)[/tex]?

   [tex] \\ [/tex]

   [tex]\huge{\underline{\textsf{\textbf{\red{Solution…}}}}}[/tex]

   Here, we are given three points as follows :

   • [tex]\green{\textsf{\textbf{A:(-6,10)}}}[/tex]
   • [tex]\green{\textsf{\textbf{B:(3,6)}}}[/tex]
   • [tex]\green{\textsf{\textbf{C:(-4,6)}}}[/tex]

   Now, we shall find in what ratio the point C divides the line joining the points A and C.

   We should find the ratio between AC and CB :

   Let the required ratio be : [tex]\green{\sf{k:1}}[/tex].

   So, now :

   • [tex]\sf m_1: k[/tex]
   • [tex]\sf m_2: 1[/tex]
   • [tex]\sf x_1: -6[/tex]
   • [tex]\sf x_2: 3[/tex]
   • [tex]\sf y_1:10[/tex]
   • [tex]\sf y_2:6[/tex]
   • [tex] \sf x:-4[/tex]
   • [tex]\sf y:6[/tex]

   Using the section formula :

   [tex]\green{:\implies{\pmb{\sf{x = \dfrac{ m_{1} x_{2} + m_{2} x_{1}}{ m_{1} + m_{2}}}}}}[/tex]

   [tex] \sf: \implies – 4 = \dfrac{k(3) + 1( – 6)}{k + 1} [/tex]

   [tex] \sf: \implies – 4 = \dfrac{3k – 6}{k + 1} [/tex]

   [tex] \sf: \implies – 4(k + 1) = 3k – 6[/tex]

   [tex] \sf: \implies – 4k – 3k = – 6 + 4[/tex]

   [tex] \sf: \implies – 7k = -2[/tex]

   [tex]\green{:\implies{\pmb{\sf{k = \dfrac{2}{7}}}}}[/tex]

   Now, substituting the value of k in k : 1 :

   [tex]\sf : \implies \dfrac{2}{7} : 1[/tex]

   We’ll multiply the ratio with 7 to get the non-fractional values :

   [tex] \sf: \implies7 (\dfrac{2}{7}) : 1(7)[/tex]

   [tex]\green{:\implies{\pmb{\sf{2 : 7}}}}[/tex]

   [tex] \\ [/tex]

   ___________________

   Therefore, the point (–4,6) divides the line segment joining the points (–6,10) and (3,6) in the ratio of 2 : 7.

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