Question

The combined equation of the straight lines of the form $y=kx+1$ (where $k$ is an integer) such that the point of intersection of each with the line $3x+4y=9$ has an integer as its $x$ - coordinate is

• A. $(y+x+1)(y+2x-1)=0$
• B. $(y+x-1)(y+2x+1)=0$
• C. $(y+x+1)(y+2x+1)=0$
• D. $(y+x-1)(y+2x-1)=0$

Answer 1

Answer: (D)

Given,
$y=kx+1...(i)$
and $3x+4y=9...(ii)$
Put the value of $y$ by Eq. (i) in Eq (ii), we get
$3x+4(kx+1)=9$
$3x+4kx+4=9$
$\Rightarrow x(3+4k)+4-9=0$
$\Rightarrow x(3+4k)-5=0$
$\Rightarrow x(3+4k)=5$
$\Rightarrow x=\frac{5}{3+4k}$
$Sx=\pm 1,\pm 5[\because x$ is an integer $]$
$\therefore 3+4k=\pm 1$ or $3+4k=\pm 5$
$k=\frac{-1}{2},-1,-2,\frac{1}{2}$
$\therefore$ we can choose value of $k$ only integer number.
$\therefore k=-1,-2$
Hence, combined equation is
$\{y=(-1)x+1\}\{y=(-2)x+1\}=0$
$\Rightarrow (y=-x+1)(y=-2x+1)=0$
$\Rightarrow (x+y-1)(2x+y-1)=0$