The correct option is D 4x+8y−7=0
Line midway between ax+by+c1=0 and ax+by+c2=0 is ax+by+c1+c22=0
Hence, line midway between 2x+4y−2=0 and 2x+4y−5=0 is 2x+4y−72=0
⇒4x+8y−7=0
Alternate Solution:
2x+4y−2=0 and 2x+4y−5=0
Let 2x+4y+k=0 be the required line.
Now
|k+2|√20=|k+5|√20⇒|k+2|=|k+5|⇒k=−72
So the line will be
⇒4x+8y−7=0