Question

The lines $x+y=10,2x-y-6=0\text{and}x+2y=k$ are concurrent if $k=$

• A. $\frac{10}{3}$
• B. $5$
• C. $\frac{44}{3}$
• D. $\frac{17}{3}$

Answer 1

The correct option is C

$\frac{44}{3}$

Since we are given the equation of lines which are given to be concurrent, lets apply the condition for concurrency for three lines.

Please note that before applying this condition we have to make sure that these lines are not parallel to each other in which case we cannot apply this condition.

From the coefficients of the equation we can see that the given lines are not parallel to each other. So now we can apply the condition,

$\left|\begin{array}{cccc}1& 1& -10& \\ 2& -1& -6& \\ 1& 2& -k& \end{array}\right|=0$

$\left|\begin{array}{cccc}1& 1& -10& \\ 0& -3& 14& \\ 0& 1& -k+10& \end{array}\right|=0$

i.e. $1(-3)(10-k)-14=0$

$\Rightarrow -30+3k-14=0$

$k=\frac{44}{3}$