`5x-4y+1=0 Rightarrow y=(5)/(4)x+(1)/(4)`
Slope of this given line is `(5)/(4)`
Now, the equatoin of any line through the intersection of the given line is of the form
`(3x+y-9)+k(4x+3y-7)=0......(i)`
`Rightarrow (3+4k)x+(1+3k)y-(9+7k)=0`
`Rightarrow (1+3k)y=-(3+4k)x+(9+7k)`
`Rightarrow y=-((3+4k))/((1+3k))x+((9+7k))/((1+3k))..............(ii)`
Let m be the slope of the line perpendicular to the required line.
`"Then,"mxx(5)/(4)-1 Rightarrow m=(-4)/(5)`
`"Then,"mxx(5)/(4)-1 Rightarrow m=(-4)/(5)`
`therefore "we must have"(-(3+4k))/((1+3k))=(-4)/(5)`
`Rightarrow 15+20k=4+12k Rightarrow 8k=-11 Rightarrow k=(-11)/(8)`
Substituting `k=(-11)/(8)(4+3y-7)=0`
`(3x+y-9)-(11)/(8)(4+3y-7)=0`
`Rightarrow (24x+8y-72)-44x-33y+77=0`
`Rightarrow 20x+25y-5=0 Rightarrow 4x+5y-1=0`
