Question

Equation of straight line midway between the lines $x+2y-1=0$ and $2x+4y-5=0$ is

• A. $2x+4y-3=0$
• B. $2x+4y-7=0$
• C. $x+2y-3=0$
• D. $4x+8y-7=0$

Answer 1

The correct option is D $4x+8y-7=0$

Line midway between $ax+by+c_1=0$ and $ax+by+c_2=0$ is $ax+by+\frac{c_1+c_2}{2}=0$
Hence, line midway between $2x+4y-2=0$ and $2x+4y-5=0$ is $2x+4y-\frac{7}{2}=0$
$\Rightarrow 4x+8y-7=0$

Alternate Solution:
$2x+4y-2=0$ and $2x+4y-5=0$
Let $2x+4y+k=0$ be the required line.
Now
$\frac{|k+2|}{\sqrt{20}}=\frac{|k+5|}{\sqrt{20}}\Rightarrow |k+2|=|k+5|\Rightarrow k=-\frac{7}{2}$
So the line will be
$\Rightarrow 4x+8y-7=0$