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Find the coordinates of the point which divides the line segment joining the points $(-2,3,5)$ and $(1,-4,6)$ in the ratio $2:3$ externally.

$\begin{array}{1 1} \big(-\large\frac{4}{5},\frac{1}{5},\frac{27}{5}\big) \\ (-8,17,3) \\ \big(\large\frac{8}{5},\frac{17}{5},\frac{27}{5}\big) \\\big(-\large\frac{8}{5},\frac{1}{5},\frac{27}{5}\big)\end{array} $

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  • Section formula: The coordinates of the point $C$ that divides the segment joining the points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ in the ratio $l:m$ externally is given by $C\big(\large\frac{lx_2-mx_1}{l-m},\frac{ly_2-my_1}{l-m},\frac{lz_2-mz_1}{l-m}\big)$
Given two points are $A(-2,3,5)$ and $B(1,-4,6)$
We know that from section formula the coordinates of the point $C$ that divides
the segment joining the points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ in the ratio $l:m$ externally is given by
$C\big(\large\frac{lx_2-mx_1}{l-m},\frac{ly_2-my_1}{l-m},\frac{lz_2-mz_1}{l-m}\big)$
$\therefore $ the coordinates of the point $C$ which divides $AB$ in the ratio $2:3$ externally is
$C\big(\large\frac{2.1-3.(-2)}{2-3},\frac{2.(-4)-3.3}{2-3},\frac{2.6-3.5}{2-3}\big)$
$=C(-8,17,3)$
answered May 5, 2014 by rvidyagovindarajan_1
edited May 5, 2014 by rvidyagovindarajan_1
 

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