Question

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) $\left( a,b \right)$

Answer 1

Hint: The standard equation of the line is represented by the formula $y=mx+c$ and it is a general equation of line. We can find the equation of line joining two given points using this and then we can find out which of the given options lie on this line.

Complete step-by-step answer:
The figure for the question by substituting a=b=1 is given by,

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.