Given the family of lines, a(2x+y+4)+b(x−2y−3)=0.Among the lines of the family, the number of lines situated at a distance of 10 from the point M(2,−3) is
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a
0
b
1
c
2
d
∞
answer is D.
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Detailed Solution
The point of intersection of the two lines are (-1, -2).Distance PM = 10Hence the required line is one which passes through (−1,−2) and is ⊥ to P.M.