Question

The lines represented by the equation ${{x}^{2}}-{{y}^{2}}-x+3y-2=0$ are
(a) $x+y-1=0,x-y+2=0$
(b) $x-y-2=0,x+y+1=0$
(c) $x+y+2=0,x-y-1=0$
(d) $x-y+1=0,x+y-2=0$

Answer 1

Hint: We need to use the completing the square method in order to contract the given equation as a difference of two squares. For this, we first need to arrange all the x and y terms together. Then, we have to add and subtract the square of the half of the coefficient of x. Then we need to divide and multiply the coefficient of x by two. Similar procedure is to be followed for the y terms. Then by using the identity ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$, we will obtain the given equation as a difference of two squares, from which the equations of the lines can be found out.

Complete step-by-step solution:
The equation given to us in the above question is written as
$\Rightarrow {{x}^{2}}-{{y}^{2}}-x+3y-2=0$
Arranging the x and the y terms together, we write the above equation as
$\Rightarrow {{x}^{2}}-x-{{y}^{2}}+3y-2=0$
We try to contract the x terms using the completing the square method. For this, we add and subtract the square of the half of the coefficient of x in the LHS of the above equation to get
$\Rightarrow {{x}^{2}}-x+{{\left( -\dfrac{1}{2} \right)}^{2}}-{{\left( -\dfrac{1}{2} \right)}^{2}}-{{y}^{2}}+3y-2=0$
Now, let us divide and multiply the coefficient of x by two to get
$\begin{align}
  & \Rightarrow {{x}^{2}}-\dfrac{2x}{2}+{{\left( -\dfrac{1}{2} \right)}^{2}}-{{\left( -\dfrac{1}{2} \right)}^{2}}-{{y}^{2}}+3y-2=0 \\
 & \Rightarrow {{x}^{2}}-2x\left( \dfrac{1}{2} \right)+{{\left( \dfrac{1}{2} \right)}^{2}}-{{\left( -\dfrac{1}{2} \right)}^{2}}-{{y}^{2}}+3y-2=0 \\
\end{align}$
Now, using the algebraic identity ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$, we can contract the first three terms as
$\begin{align}
  & \Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}-{{\left( -\dfrac{1}{2} \right)}^{2}}-{{y}^{2}}+3y-2=0 \\
 & \Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}-\dfrac{1}{4}-{{y}^{2}}+3y-2=0 \\
 & \Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}-{{y}^{2}}+3y-2-\dfrac{1}{4}=0 \\
 & \Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}-\left( {{y}^{2}}-3y \right)-2-\dfrac{1}{4}=0 \\
\end{align}$
Following the similar procedure, we add and subtract ${{\left( -\dfrac{3}{2} \right)}^{2}}$ inside the bracket to get
\[\begin{align}
  & \Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}-\left( {{y}^{2}}-3y+{{\left( -\dfrac{3}{2} \right)}^{2}}-{{\left( -\dfrac{3}{2} \right)}^{2}} \right)-2-\dfrac{1}{4}=0 \\
 & \Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}-\left( {{y}^{2}}-3y+{{\left( \dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}} \right)-2-\dfrac{1}{4}=0 \\
\end{align}\]
Now, we divide and multiply the coefficient of y by two to get
\[\Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}-\left( {{y}^{2}}-2\left( \dfrac{3}{2} \right)y+{{\left( \dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}} \right)-2-\dfrac{1}{4}=0\]
Now, using the algebraic identity ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$ we can wrie
\[\begin{align}
  & \Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}-\left( {{\left( y-\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}} \right)-2-\dfrac{1}{4}=0 \\
 & \Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}-{{\left( y-\dfrac{3}{2} \right)}^{2}}+{{\left( \dfrac{3}{2} \right)}^{2}}-2-\dfrac{1}{4}=0 \\
\end{align}\]
Opening the square, we get
\[\begin{align}
  & \Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}-{{\left( y-\dfrac{3}{2} \right)}^{2}}+\dfrac{9}{4}-2-\dfrac{1}{4}=0 \\
 & \Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}-{{\left( y-\dfrac{3}{2} \right)}^{2}}+\dfrac{8}{4}-2=0 \\
 & \Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}-{{\left( y-\dfrac{3}{2} \right)}^{2}}+2-2=0 \\
 & \Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}-{{\left( y-\dfrac{3}{2} \right)}^{2}}=0 \\
\end{align}\]
Adding \[{{\left( y-\dfrac{3}{2} \right)}^{2}}\] both the sides, we get
\[\begin{align}
  & \Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}={{\left( y-\dfrac{3}{2} \right)}^{2}} \\
 & \Rightarrow {{\left( y-\dfrac{3}{2} \right)}^{2}}={{\left( x-\dfrac{1}{2} \right)}^{2}} \\
 & \Rightarrow y-\dfrac{3}{2}=\pm \left( x-\dfrac{1}{2} \right) \\
\end{align}\]
Case I:
\[\Rightarrow y-\dfrac{3}{2}=x-\dfrac{1}{2}\]
Adding $\dfrac{3}{2}$ both the sides, we get
$\begin{align}
  & \Rightarrow y=x+\dfrac{3}{2}-\dfrac{1}{2} \\
 & \Rightarrow y=x+\dfrac{2}{2} \\
 & \Rightarrow y=x+1 \\
\end{align}$
Subtracting $y$ from both sides, we get
$\begin{align}
  & \Rightarrow 0=x-y+1 \\
 & \Rightarrow x-y+1=0 \\
\end{align}$
Case II:
$\Rightarrow y-\dfrac{3}{2}=-\left( x-\dfrac{1}{2} \right)$
Adding $\dfrac{3}{2}$ both the sides, we get
$\begin{align}
  & \Rightarrow y=-x+\dfrac{3}{2}+\dfrac{1}{2} \\
 & \Rightarrow y=-x+\dfrac{4}{2} \\
 & \Rightarrow y=-x+2 \\
\end{align}$
Adding $x$ both sides, we get
$\begin{align}
  & \Rightarrow y+x-2=0 \\
 & \Rightarrow x+y-2=0 \\
\end{align}$
Thus, the equations of lines we got are $x-y+1=0$ and $x+y-2=0$.
Hence, the correct answer is option (d).

Note: Make sure that you are left with only two square terms in the end after the simplification of the given equation. This is because if there is any residual constant term along with the square terms, then we cannot obtain the equation of the lines from that equation. So if you got any residual constant term, that means there is some error in your simplification.