CBSE Class 10 Answered
aruthmetic progressions
Asked by | 28 Jan, 2009, 07:47: AM
Expert Answer
a,b,c are in A.P.
So,
2b=a+c
So,
b=(a+c)/2
So, the equation of the line ax+by+c=0
can be written as
ax+(a+c/2)y+c=0
So we get,
2ax+(a+c)y+2c=0
Now if (1,-2) has to lie on this line, we must have that the coordinates of this point always satisfy the equation .
So, putting x=1, y= -2 in the equation of the line we get,
2a-2(a+c)+2c
=2(a+c)-2(a+c)
=0
So we see that the point (1,-2) satisfies the equation of the line whatever may be the values of a,b,c as long as they are in A.P.
Hence proved.
Answered by | 28 Jan, 2009, 10:10: AM
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