Two points $\left( a,0 \right)$ and $\left( 0,b \right)$ are joined by a straight line. Another point on this line is
(a) \[\left( 3a,-2b \right)\]
(b) $({{a}^{2}},ab)$
(c) \[\left( -3a,2b \right)\]
(d) $\left( a,b \right)$

Now we the equation of the line is represented by the formula $y=mx+c$
Since, two points $\left( a,0 \right)$ and $\left( 0,b \right)$ are joined by a straight line this implies that we can substitute these points in the equation. We get $\left( a,0 \right)$ on the line $y=mx+c$ implies $0=ma+c$ which is further equal to $ma=-c$ and hence thus the value of $m=\dfrac{-c}{a}$
Also, $\left( 0,b \right)$ on the line $y=mx+c$ implies $b=m0+c$ and we get $b=c$ so now we substitute the values of $m$ and $c$ in the standard equation of straight line $y=mx+c$
$\begin{align}
  & \Rightarrow y=\left( \dfrac{-c}{a} \right)x+b \\
 & \Rightarrow y=\dfrac{-cx}{a}+b
\end{align}$
We can write this equation as $ay=-cx+ab...(i)$ and now we substitute given options one by one. First we consider \[\left( 3a,-2b \right)\] in equation $(i)$ and so $ay=-cx+ab$
$\begin{align}
  & \Rightarrow a\left( -2b \right)=-c\left( 3a \right)+ab \\
 & \Rightarrow 3ac=ab+2ab \\
 & \Rightarrow 3ac=3ab \\
 & \Rightarrow c=b \\
\end{align}$
Now we consider the point $({{a}^{2}},ab)$ in equation $(i)$ and so the equation $ay=-cx+ab$
$\begin{align}
  & \Rightarrow a\left( ab \right)=-c\left( {{a}^{2}} \right)+ab \\
 & \Rightarrow {{a}^{2}}b=-c{{a}^{2}}+ab \\
 & \Rightarrow {{a}^{2}}b+c{{a}^{2}}=ab \\
 & \Rightarrow {{a}^{2}}\left( b+c \right)=ab \\
 & \Rightarrow a(b+c)=b \\
\end{align}$
Now we substitute the point \[\left( -3a,2b \right)\] in equation $(i)$ and the equation $ay=-cx+ab$
$\begin{align}
  & \Rightarrow a\left( 2b \right)=-c\left( -3a \right)+ab \\
 & \Rightarrow 2ab=3ac+ab \\
 & \Rightarrow -3ac=ab-2ab \\
 & \Rightarrow -3ac=-ab \\
 & \Rightarrow 3ac=b \\
\end{align}$
Lastly we substitute the value $\left( a,b \right)$ in equation $(i)$ and so we have,
$\begin{align}
  & ay=-cx+ab \\
 & \Rightarrow ab=-ca+ab \\
 & \Rightarrow 0=-ca \\
\end{align}$
Now we see that it is clear that only the\[\left( 3a,-2b \right)\] point satisfies the equation out of the other three. Hence the right option is (a)

Note: Alternatively we could have used section formula $\left( \dfrac{{{m}_{1}}{{x}_{2}}+{{m}_{2}}{{x}_{1}}}{{{m}_{2}}+{{m}_{1}}},\dfrac{{{m}_{1}}{{y}_{2}}+{{m}_{2}}{{y}_{1}}}{{{m}_{2}}+{{m}_{1}}} \right)$ in this question but here we are not given any type of ratio. Be careful while shifting values on either side of the equal sign. If there is a negative sign which has been shifted on the other side of equal will become positive and vice-versa.