Geo2.41

Find the value of k if the lines $2x+3y=6,kx-9y+4=0,5x+6y=13\,$ are concurrent.

Solving the given first and third equations

$5(2x+3y-6)-2(5x+6y-13)=0\,$

$10x+15y-30-10x-12y+26=0\,$

$3y-4=0,y=\frac{4}{3}\,$

Now the value of x is $2x+3(\frac{4}{3}=6,2x=2,x=1\,$

The point of intersection of the three lines is $\left(1,\frac{4}{3}\right)\,$

This point passes thro' the second line,hence

$k(1)-9(\frac{4}{3})+4=0,k-12+4=0,k=8\,$

Hence the value of k is 8.