State whether the following statements are true or false. Justify your answer.
Points $ A(-6,10), B(-4,6) $ and $ C(3,-8) $ are collinear such that $ A B=\frac{2}{9} A C $.
Given:
Points \( A(-6,10), B(-4,6) \) and \( C(3,-8) \) are collinear such that \( A B=\frac{2}{9} A C \).
To do:
We have to find whether the given statement is true or false.
Solution:
We know that,
If the area of the 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.
We know that,
Area of triangle $=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Here,
$x_{1}=-6, x_{2}=-4, x_{3}=3$ and $y_{1}=10, y_{2}=6, y_{3}=-8$
Therefore,
Area of the triangle formed by the points \( A(-6,10), B(-4,6) \) and \( C(3,-8) \) is,
Area $=\frac{1}{2}[-6\{6-(-8)\}+(-4)(-8-10)+3(10-6)]$
$=\frac{1}{2}[-6(14)+(-4)(-18)+3(4)]$
$=\frac{1}{2}(-84+72+12)$
$=0$
This implies,
The given points are collinear.
The distance between $A(-6,10)$ and $B(-4,6)$ is,
$A B=\sqrt{(-4+6)^{2}+(6-10)^{2}}$
$=\sqrt{2^{2}+4^{2}}$
$=\sqrt{4+16}$
$=\sqrt{20}$
$=2 \sqrt{5}$
The distance between $A(-6,10)$ and $C(3,-8)$ is,
$A C=\sqrt{(3+6)^{2}+(-8-10)^{2}}$
$=\sqrt{9^{2}+18^{2}}$
$=\sqrt{81+324}$
$=\sqrt{405}$
$=\sqrt{81 \times 5}$
$=9 \sqrt{5}$
$A B=\frac{2}{9} A C$
Hence, the given statement is true.
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