1.

Let L be the line belonging to the family of straight lines (a+2b) x+(a-3b)y+a-8b =0, a, b in R, which is the farthest from the point (2, 2). If L is concurrent with the lines x-2y+1=0 and 3x-4y+ lambda=0, then the value of lambda is

Answer»

2
1
-4
5

Solution :LINES x+4y+7 =0 and x-2y+1 =0 INTERSECT at (-3, -1) which must satisfy the line `3x-4y+lambda=0. "Then " -9+4+lambda=0 " or " lambda = 5.`


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