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Using the section formula show that the points $A(2,-3,4),\:B(-1,2,1),\: C(0,\large\frac{1}{3}$$,2)$ are collinear.

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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 $l:m$ internally 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 points are $A(2,-3,4),\:B(-1,2,1),\: C(0,\large\frac{1}{3}$$,2)$
If the three points are collinear then according to section formula,
$C$ divides $AB$ in in some ratio.
Let the ratio in which $C$ divides $AB$ be $k:1$
We know that from section formula,
the coordinates of $C$ is given by $C\big(\large\frac{-k+2}{k+1},\frac{2k-3}{k+1},\frac{k+4}{k+1}\big)$
$\therefore$ By comparing the given coordinates of $C(0,\large\frac{1}{3}$$,2)$
we get $\large\frac{-k+2}{k+1}$$=0$, $\large\frac{2k-3}{k+1}=\frac{1}{3}$ and $\large\frac{k+4}{k+1}$$=2$
$\Rightarrow\:k=2$ in all the three equations.
$\therefore$ The given three points are collinear.
answered May 8, 2014 by rvidyagovindarajan_1

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