Question

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

Answer 1

True
If the area of triangle formed by the points $(x_1,y_1),(x_2,y_2)and(x_3,y_3)$ is zero, then the points are collinear.
$\because \text{Area of triangle}=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]Here,x_1=-6,x_2=-4,x_3=3and{y}_1=10,y_2=6,y_3=-8\because Areaof\Delta ABC=\frac{1}{2}[-6(6-(-8))+(-4\left)\right(-8-10)+3(10-6\left)\right]=\frac{1}{2}[-6(14)+(-4\left)\right(-18)+3(4\left)\right]=\frac{1}{2}[-84+72+12]=0\text{So, given points are collinear.}\text{Now, distance between A(-6,10) and B(-4,6).}AB=\sqrt{(-4+6{)}^2+(6-10{)}^2}=\sqrt{2^2+4^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$

$\left[\begin{array}{c}\because \text{Distance (d) between the points}\\ (x_1,y_1)and(x_2,y_2),\\ d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}\end{array}\right]$

$\text{Distance between A(-6,10) and C(3,-8),}AC=\sqrt{(3+6{)}^2+(-8-10{)}^2}=\sqrt{9^2+{18}^2}=\sqrt{81+324}=\sqrt{405}=\sqrt{81\times 5}=9\sqrt{5}\therefore AB=\frac{2}{9}AC\text{which is the required relation.}$