Question

If the lines $x+\left( a-1 \right)y=1$ and $2x+{{a}^{2}}y=1$ where $a\in R-\left\{ 0,1 \right\}$ are perpendicular to each other. Then distance of their points of intersection from the origin is\[\]
A.$\dfrac{5}{2}$\[\]
B.$\dfrac{2}{\sqrt{5}}$\[\]
C.$\dfrac{\sqrt{5}}{2}$\[\]
D. $\sqrt{\dfrac{2}{5}}$\[\]

Answer 1

Hint: We convert both the equations to the slope point form $y=mx+c$. We use the obtained slopes ${{m}_{1}},{{m}_{2}}$ in the identity of perpendicular lines ${{m}_{1}}{{m}_{2}}=-1$ . We solve the equation to get the value of $a$. We put $a$ in the two given equations and solve the resulting linear of pair of equation to the co-ordinate of point of intersection. We use the formula $d=\sqrt{{{x}_{1}}^{2}+{{y}_{1}}^{2}}$ for any point $\left( {{x}_{1}},{{y}_{1}} \right)$ to find the distance from the origin. \[\]

 We know from the factor theorem that a polynomial $p\left( x \right)$ has a factor $\left( x-a \right)$ if and only if $p\left( a \right)=0$ in other words $a$ is a zero of $p\left( x \right)$ .We also know that the quadratic equation $a{{x}^{2}}+bx+c=0,a\ne 0$ has real root if the discriminant $D={{b}^{2}}-4ac<0$.\[\]

We know that the equation of lines with $m$ and $y-$intercept is given by (called slope-intercept form)
\[y=mx+c\]

Complete step by step answer:
The given equation of two lines are
\[\begin{align}
  & x+\left( a-1 \right)y=1\Rightarrow y=\dfrac{-1}{a-1}x+1 \\
 & 2x+{{a}^{2}}y=1\Rightarrow y=\dfrac{-2}{{{a}^{2}}}x+1 \\
\end{align}\]
We convert the equations to the slope-point form and get,
\[\begin{align}
  & x+\left( a-1 \right)y=1...(1) \\
 & 2x+{{a}^{2}}y=1...(2) \\
\end{align}\]
So we have slope of the line(1) say ${{m}_{1}}=\dfrac{-1}{a-1}$ and slope of line(2) say ${{m}_{2}}=\dfrac{-2}{{{a}^{2}}}$.We know that when two lines are perpendicular the product of their slope id $-1$. So we have,
\[\begin{align}
  & \Rightarrow {{m}_{1}}{{m}_{2}}=-1 \\
 & \Rightarrow \dfrac{-1}{a-1}\times \dfrac{-2}{{{a}^{2}}}=-1 \\
 & \Rightarrow 2=-{{a}^{2}}\left( a-1 \right) \\
 & \Rightarrow {{a}^{2}}\left( a-1 \right)+2=0 \\
 & \Rightarrow {{a}^{3}}-a+2=0 \\
\end{align}\]
We want to find the of the above cubic equation . We put the value $a=-1$ and finds that it zero of the polynomial ${{a}^{3}}-a+2$. We use the factor theorem to conclude that that $a-\left( -1 \right)=a+1$ will be a factor of ${{a}^{3}}-a+2$. We divide ${{a}^{3}}-a+2$ by $a+1$ using long division method and get the other factor ${{a}^{2}}-2a+2$. So we have
\[\begin{align}
  & \Rightarrow \left( a+1 \right)\left( {{a}^{2}}-a+2 \right)=0 \\
 & \Rightarrow a+1=0\text{ or }{{a}^{2}}-a+2=0 \\
\end{align}\]
So we have one real root as $a=-1$. When we go to find the other roots in ${{a}^{2}}-2a+2=0$ we find the discriminant of the quadratic equation $D={{\left( -2 \right)}^{2}}-4\times 2\times 1=-4$ less than 0. So we will not get any real root from ${{a}^{2}}-2a+2=0$. So the only real value of $a$ is $-1.$ We put $a=-1$ in equation (1) ,(2) and get
\[\begin{align}
  & x-2y=1 \\
 & 2x+y=1 \\
\end{align}\]
We solve the above pair of linear equations and get
\[x=\dfrac{-1}{5},y=\dfrac{3}{5}\]
So we now have the co-ordinate of point of intersection of line(1) and line(2) say $P\left( \dfrac{-1}{5},\dfrac{3}{5} \right)$. So the distance from the origin is
\[OP=\sqrt{{{\left( \dfrac{-1}{5} \right)}^{2}}+{{\left( \dfrac{3}{5} \right)}^{2}}}=\sqrt{\dfrac{10}{25}}=\sqrt{\dfrac{2}{5}}\]

So the correct option is D.\[\]

Note:
The slopes ${{m}_{1}}=\dfrac{-1}{a-1}$ and ${{m}_{2}}=\dfrac{-2}{{{a}^{2}}}$ exist because we are given the condition $a\in R-\left\{ 0,1 \right\}$. The formula for distance from the origin $d=\sqrt{{{x}_{1}}^{2}+{{y}_{1}}^{2}}$ is only a special case of the distance between two points say $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ which is $d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$.