Question

The joint equation of the straight lines $x+y=1$ and $x-y=4$ is
A. ${{x}^{2}}-{{y}^{2}}=-4$
B. ${{x}^{2}}-{{y}^{2}}=4$
C. $(x+y-1)(x-y-4)=0$
D. $(x+y+1)(x-y+4)=0$

Answer 1

Hint: In this equation, we are to find the combined form of the given two lines. For this, we need to multiply the given lines and by simplifying them, we get the required equation.



Formula Used:The combined equation of pair of straight lines is written as
$H\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}=0$
This is called a homogenous equation of the second degree in $x$ and $y$
And
$S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
This is called a general equation of the second degree in $x$ and $y$.
If ${{h}^{2}}If ${{h}^{2}}=ab$, then $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents coincident lines.
If ${{h}^{2}}>ab$, then $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents two real and different lines that pass through the origin.
Thus, the equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents two lines. They are:
$ax+hy\pm y\sqrt{{{h}^{2}}-ab}=0$



Complete step by step solution:Given lines are $x+y=1$ and $x-y=4$
Then, for the combined equation of these two lines, we get
$(x+y-1)(x-y-4)=0\text{ }...(1)$
On simplifying the above equation, we get
$\begin{align}
  & (x+y-1)(x-y-4)=0 \\
 & \Rightarrow {{x}^{2}}-xy-4x+xy-{{y}^{2}}-4y-x+y+4=0 \\
 & \Rightarrow {{x}^{2}}-{{y}^{2}}-5x+y+4=0 \\
\end{align}$
But it in the given options, we have only the product form of the given lines. So, the combined form of the given two lines is $(x+y-1)(x-y-4)=0$.



Option ‘C’ is correct



Note: Here, in this question, the combined equation of the given two lines $x+y=1$ and $x-y=4$, are multiplied by writing them in the form of $ax+by+c=0$. Since we know that, a product of any two factors gives an equation, the product of these two lines gives another equation that is said to be the combined equation of these two lines.