Question

Prove that the points A(3a,0),B(0,3b),C(a,2b) are collinear

Answer 1

Answer:

substitute the given value in area of triangle formula and equate it to 0

Step-by-step explanation:

Answer 2

Answer:

Given Points are collinear.

Step-by-step explanation:

Given Points are ( 3a , 0 ) , ( 0 , 3b ) & ( a , 2b )

To prove points are collinear

We use formula of Area of triangle by point.

If Area of triangle equals to 0 then points are collinear.

Area of triangle is given by,

$Area\:of\:triangle=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

$Area\:of\:triangle=\frac{1}{2}\left|3a(3b-2b)+0(2b-0)+a(0-3b)\right|$

$Area\:of\:triangle=\frac{1}{2}\left|3ab-3ab\right|$

$Area\:of\:triangle=0$

Therefore, Given Points are collinear.