If $a ≠b ≠0$, prove that the points $(a, a^2), (b, b^2), (0, 0)$ are never collinear.
Given:
Given points are $(a, a^2), (b, b^2), (0, 0)$.
$a ≠ b ≠ 0$
To do:
We have to prove that the given points are never collinear.
Solution:
Let $A(a, a^2), B(b, b^2)$ and $C(0, 0)$ be the vertices of $\triangle ABC$.
We know that,
If the points $A, B$ and $C$ are collinear then the area of $\triangle ABC$ is zero.
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle \( ABC=\frac{1}{2}[a(b^2-0)+b(0-a^2)+0(a^2-b^2)] \)
\( =\frac{1}{2}[ab^2-a^2b+0] \)
\( =\frac{1}{2}[ab(b-a)] \)
\( ≠0 \) (Since $a ≠ b ≠ 0$)
Here,
The area of $\triangle ABC$ is not equal to zero.
Therefore, points $A, B$ and $C$ are not collinear.
Hence proved.  
Related Articles
- If points $( a,\ 0),\ ( 0,\ b)$ and $( x,\ y)$ are collinear, prove that $\frac{x}{a}+\frac{y}{b}=1$.
- Prove that the points $(a, 0), (0, b)$ and $(1, 1)$ are collinear if, $\frac{1}{a} + \frac{1}{b} = 1$.
- If $a ≠b ≠c$, prove that the points $(a, a^2), (b, b^2), (c, c^2)$ can never be collinear.
- Find the distance between the points $A( 0,\ 6)$ and $B ( 0,\ -2)$.
- If \( x=\frac{\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}+b^{2}}-\sqrt{a^{2}-b^{2}}} \), then prove that \( b^{2} x^{2}-2 a^{2} x+b^{2}=0 \).
- If the roots of the equation $(c^2-ab)x^2-2(a^2-bc)x+b^2-ac=0$ are equal, prove that either $a=0$ or $a^3+b^3+c^3=3abc$.
- If \( a^{2}-b^{2}=0 \), then the value of \( \frac{a}{b} \) is____.
- If the roots of the equations $ax^2+2bx+c=0$ and $bx^2-2\sqrt{ac}x+b=0$ are simultaneously real, then prove that $b^2=ac$.
- Prove that the point $(-2, 5), (0, 1)$ and $(2, -3)$ are collinear.
- If the roots of the equation $(b-c)x^2+(c-a)x+(a-b)=0$ are equal. Prove that $2b=a+c$
- Show that $\triangle ABC$, where $A (-2, 0), B (2, 0), C (0, 2)$ and $\triangle PQR$, where $P (-4, 0), Q (4, 0), R (0, 4)$ are similar.
- If the roots of the equation $a(b-c) x^2+b(c-a) x+c(a-b) =0$ are equal, then prove that $b(a+c) =2ac$.
- If the roots of the equation $(b-c)x^2+(c-a)x+(a-b)=0$ are equal, then prove that $2b=a+c$.
- If the roots of the equation $(a^2+b^2)x^2-2(ac+bd)x+(c^2+d^2)=0$ are equal, prove that $\frac{a}{b}=\frac{c}{d}$.
- Are the following pair of linear equations consistent? Justify your answer.\( 2 a x+b y=a \)\( 4 a x+2 b y-2 a=0 ; \quad a, b≠0 \)
Kickstart Your Career
Get certified by completing the course
Get Started