if the straight line ax+by+c =0 always passes through (1,-2) then a,b,c are in
1.ap
2.gp
3. hp
4. none of these
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Answered by
2
Δ Given that a straight line ax+by+c = 0 is always passe through the point (1,-2)
hence the value of x = 1 and y =-2
therefore from Eqn ,
a(1)+b(-2)+c = 0
⇒ a-2b+c = 0
⇒a+c = 2b
so, b= (a+c)/2
therefore a, b ,c are in A.P
hence the value of x = 1 and y =-2
therefore from Eqn ,
a(1)+b(-2)+c = 0
⇒ a-2b+c = 0
⇒a+c = 2b
so, b= (a+c)/2
therefore a, b ,c are in A.P
Answered by
0
Given the straight line ax+by+c=0 always passes through (1,-2)
Hence x=1 and y=-2
now, a(1)+b(-2)+c=0
a-2b+c=0
a+c=2b
Hence a, b, c are in A.P
Hence x=1 and y=-2
now, a(1)+b(-2)+c=0
a-2b+c=0
a+c=2b
Hence a, b, c are in A.P
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