Question

If the lines $2 x+y-3=0,5 x+K y-3=0$ and $3 x-y-2=0$ 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
$2 x+y-3=0 .....$(i)
$5 x+k y-3=0 .....$(ii)
$3 x-y-2=0....$(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
$5.1+k \cdot 1-3=0 \text { or } k=-2$