If a b 0, prove that the points (a, a^2 ), (b, b^2 ),(0,0) will not be collinear.

Let A ( a , a ^2 ), B ( b , b ^2 ) and C (0,0) be the coordinates of the given points. We know that the area of triangle having vertices (x_1, y_1 ), (x_2, y_2 ) and (x_3, y_3 ) is |1 2 [ x _1 ( y _2- y _3 )+ x _2 ( y _3- y _1 )+ x _3 ( y _1- y _2 ) ] | square units. So, Area of ABC = |1 2 [ a ( b ^2-0 )+ b (0- a ^2 )+0 ( a ^2- b ^2 ) ] | = |1 2 (a b^2-a^2 b ) | =1 2 | ab ( b - a )| 0( a b 0) Since the area of the triangle formed by the points (a, a^2 ), (b, b^2 ) and (0,0) is not zero, so the given points are not collinear.

If a b 0, prove that the points (a, a^2 ), (b, b^2 ),(0,0) will n…

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Question

If $a \neq b \neq 0$, prove that the points $\left(a, a^2\right),\left(b, b^2\right),(0,0)$ will not be collinear.

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Answer

Let $A \left( a , a ^2\right), B \left( b , b ^2\right)$ and $C (0,0)$ be the coordinates of the given points.
We know that the area of triangle having vertices $\left(x_1, y_1\right),\left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ is $\left|\frac{1}{2}\left[ x _1\left( y _2- y _3\right)+ x _2\left( y _3- y _1\right)+ x _3\left( y _1- y _2\right)\right]\right|$ square units.
So, Area of $\triangle ABC =\left|\frac{1}{2}\left[ a \left( b ^2-0\right)+ b \left(0- a ^2\right)+0\left( a ^2- b ^2\right)\right]\right|$ $=\left|\frac{1}{2}\left(a b^2-a^2 b\right)\right|$
$=\frac{1}{2}| ab ( b - a )|$
$\neq 0(\because a \neq b \neq 0)$
Since the area of the triangle formed by the points $\left(a, a^2\right),\left(b, b^2\right)$ and $(0,0)$ is not zero, so the given points are not collinear.