Question

If the lines $2x+y-3=0,5x+Ky-3=0$ and $3x-y-2=0$ are concurrent, then the value of $K$ is

• A. $-1$
• B. $-2$
• C. $-3$
• D. $-4$

Answer 1

Answer: (B)

Three lines are said to be concurrent, if they pass through a common point, i.e., point of intersection of any two lines lies on the third line. Here given lines are
$2x+y-3=\mathrm{0.....}$(i)
$5x+ky-3=\mathrm{0.....}$(ii)
$3x-y-2=\mathrm{0....}$(iii)
Solving Eqs. (i) and (iii) by cross-multiplication method, we get
$\frac{x}{-2-3}=\frac{y}{-9+4}=\frac{1}{-2-3}\text{ or }x=1,y=1$
Therefore, the point of intersection of two lines is $(1,1)$. Since, above three lines are concurrent, the point $(1,1)$ will satisfy Eq. (ii), so that
$5.1+k\cdot 1-3=0\text{ or }k=-2$