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 $.

AcademicMathematicsNCERTClass 10

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.

raja
Updated on 10-Oct-2022 13:28:28

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