Find the equation of a line perpendicular to the line `(x)/(a)+(y)/(b)-1` and passes through the mid-point of the line segment lying between the axes of the given line.

Explanation

Given straight line is
`(x)/(a)+(y)/(b)=1`
This line cuts the axes at points `A(a,0)` and `B(0,b)`.
Co-ordinates of mid point of `AB=((a+0)/(2),(0+b)/(2))`
`=((a)/(2),(b)/(2))`
Now, slope of given line
`m_(1)=-("Coefficient of " x)/("Coefficient of" y)=-(1//a)/(1//b)=-(b)/(a)`
Slope of perpendicular line
`m_(2)=-(1)/(m_(1))=-(1)/(b//a)=(a)/(b)`
`:.` Equation of line passes through `((a)/(2),(b)/(2))` and whose slope is `(a)/(b)` is
`y-(b)/(2)=(a)/(b)(x-(a)/(2))`
`implies 2by-b^(2)=2ax-a^(2)`
`implies2ax-2by=a^(2)-b^(2)`.