Find the value of k for which the lines 3x+y=2,k+2y=3 and 2x−y=3 may intersect at a point.
The given lines are :
On solving (i) and (iii)by cross multiplication, we get
⇒x=(−5−5)=1 and y=(5−5)=−1
Thus, the point of intersection of (i) and (iii) is P(1, -1)
For the given lines to intersect at a point, x=1 and y=−1 must satisfy (ii) also.
Hence, k = 5