Find the equation of a line passing through the point of intersection of lines `x-y-1=0` and `2x-3y+1=0` and perpendicular to the line `x-2y+5=0`.

Explanation

Let the equation of a line through the point of intersection of the lines `x-y-1=0` and `2x-3y-1=0` be
`(x-y-1)+lambda(2x-3y+1)=0`
`implies x(1+2lambda)-y(1+3lambda)+(lambda-1)=0`…..`(1)`
Its slope is `m_(1)=(1+2lambda)/(1+3lambda)`
Slope of given line `x-2y+5=0` is
`m_(2)=(1)/(2)`
For perpendicular lines
`m_(1)m_(2)=-1`
`(1+2lambda)/(1+3lambda)*(1)/(2)=-1`
`implies 1+2lambda=-2-6lambda`
`implies lambda=-(3)/(8)`
Put in `lambda=-(3)/(8)` in eq. `(1)`
`x(1-(6)/(8))-y(1-(9)/(8))+(-(3)/(8)-1)=0`
`2x+y-11=0`