The points A(-6,10), B(-4,6) and


The points $A(-6,10), B(-4,6)$ and $C(3,-8)$ are collinear such that $\mathrm{AB}=\frac{2}{9} \mathrm{AC}$.



If the area of triangle formed by the points (x1,y2), (x2, y2) and (x3, y3) is zero, then the points are collinear,

$\because \quad$ Area of triangle $=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

Here, $x_{1}=-6, x_{2}=-4, x_{3}=3$ and $y_{1}=10, y_{2}=6, y_{3}=-8$

$\therefore \quad$ Area of $\triangle A B C=\frac{1}{2}[-6\{6-(-8)\}+(-4)(-8-10)+3(10-6)]$




So, given points are collinear.

Now, distance between $A(-6,10)$ and $B(-4,6), A B=\sqrt{(-4+6)^{2}+(6-10)^{2}}$

$=\sqrt{2^{2}+4^{2}}=\sqrt{4+16}=\sqrt{20}=2 \sqrt{5}$

$\left[\because\right.$ distance between the points $\left(x_{1}, y_{1}\right)$ and $\left.\left(x_{2}, y_{2}\right), d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]$

Distance between $A(-6,10)$ and $C(3,-8), A C=\sqrt{(3+6)^{2}+(-8-10)^{2}}$


$=\sqrt{405}=\sqrt{81 \times 5}=9 \sqrt{5}$

$\therefore$ $A B=\frac{2}{9} A C$

which is the required relation.

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now