29ya me too getting only one answer...
my approach..put y = mx + c in the second degree eqn...
a quadratic eqn of x is formed..since it cut the pair of lines at only one real point so b2 = 4ac..or say Discriminant of quadratic eqn is zero..
y = mx+2 will cut the pair of lines 2x2 - 3xy + y2 - x + y = 0 , at only one real point if m is equal to
A) 1 B) -1 C) 2 D) -2
[multiple options correct]
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6 Answers
29
12x2 - 3xy + y2 - x + y = 0
(2x - y - 1)(x - y) = 0
Both r non parallel lines.
L1 is x-y=0
L2 is 2x - y - 1 = 0
L2 slope = 2, L1 slope is 1.
L1 , L2 meet at point P(1,1)
so , 1 =m + 2
m= -1
Already m = 2 and 1,
23manmay is right na , 1,-1,2 , these 3 shud be the ans .........
i.e A,B,C ....
23avinav wat u calculated is the line wich cuts both the lines ...
and the lines wich cut only one of the 2 pair of lines hav slopes = 2, 1
so -1 , 1 ,2 shud be d ans right ?
