Question

If the systems of linear equations
\[\begin{align}
  & x-2y+kz=1 \\
 & 2x+y+z=2 \\
 & 3x-y-kz=3 \\
\end{align}\]
Has a solution \[\left( x,y,z \right),z\ne 0\], then \[\left( x,y \right)\] lies on the straight line whose equation is:
$\begin{align}
  & \left( a \right)\text{ }3x-4y-1=0 \\
 & \left( b \right)\text{ }3x-4y-4=0 \\
 & \left( c \right)\text{ 4}x-3y-4=0 \\
 & \left( d \right)\text{ 4}x-3y-1=0 \\
\end{align}$

Answer 1

Hint: We have three linear equations given in the question. To get the equation of the straight line, we have to try to solve any of these equations such that we can get an equation in x and y. So, we can start by considering first and third equations and adding them. This will get us the answer.

Complete step by step answer:
Let us learn about the concept first.
Linear equations are those equations that are of the first order. These equations are defined for lines in the coordinate system. Linear equations are first-degree equations as it has the highest exponent of variables as 1. We must choose a set of 2 equations to find the values of 2 variables. Such as \[ax+by+c=0\] and \[dx+ey+f=0\], also called a system of equations with two variables, where x and y are two variables and a, b, c, d, e, f are constants, and a, b, d and e are not zero. On solving these two equations, we will get values of variable x and y. Else, the single equation has an infinite number of solutions.
Let us number the given systems of given linear equations
\[x-2y+kz=1\] … (i)
\[2x+y+z=2\] …(ii)
\[3x-y-kz=3\] …(iii)
Therefore, to find locus of (x, y) we can add equation (i) and (iii),
\[x-2y+kz+3x-y-kz=1+3\]
On simplification, we have
\[4x-3y=4\]
We can rewrite the above equation as,
\[4x-3y-4=0\]
Therefore \[\left( x,y \right)\] lies on the straight line whose equation is: \[4x-3y-4=0\].

So, the correct answer is “Option A”.

Note: Students get confused while solving this type of problem like, which two equations have to be solved. But, here in the given three equations, two equations are having constant k. So, one can get the answer, only if they try to solve equations having constant k. All the options are very similar and so special care must be taken while selecting the right one.