#### Points A (–6, 10), B (–4, 6) and C (3, –8) are collinear such that AB =2/9 AC.

Solution.     If the points A (–6, 10), B(–4, 6) and C(3, –8) are collinear then area of $\triangle$ABC = 0
$Area \, of \, \triangle ABC= \frac{1}{2}\left [ -6\left ( 6+8 \right ) +\left ( -4 \right )\left ( -8-10 \right )+3\left ( 10-6 \right )\right ]$
$= \frac{1}{2}\left [ -84+72+12 \right ]$
Area of $\triangle$ABC = 0
Hence, A, B and C are collinear
$Distance \, between \, AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$

$AB= \sqrt{\left ( -4+6 \right )^{2}+\left ( 6-10 \right )^{2}}$

$AB=\sqrt{4+16}= \sqrt{20}= 2\sqrt{5}$
$Distance \, between \, AC= \sqrt{\left ( 3+6 \right )^{2}+\left ( -8-10 \right )^{2}}$

$= \sqrt{81+324}$
$= \sqrt{405}$

$AC= 9\sqrt{5}$
Hence,
$\frac{AB}{AC}= \frac{2\sqrt{5}}{9\sqrt{5}}$
$AB= \frac{2}{9}AC$