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Question

Find the value of k for which the lines 3x+y=2,k+2y=3 and 2xy=3 may intersect at a point.


Solution

The given lines are :

3x+y2=0......(i)

kx+2y3=0......(ii)

2xy3=0......(iii)

On solving (i) and (iii)by cross multiplication, we get

x(32)=y(4+9)=1(32)x5=y5=15

x=(55)=1 and y=(55)=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.

(k×1)+2×(1)3=0k5=0k=5

Hence, k = 5

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