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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



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.

Main Page:Geometry:Straight Lines-I