The correct option is D 4x−3y+2=0
Equation of line passing through the intersection of the lines x+2y−5=0 and 3x+7y−17=0 is
(x+2y−5)+λ(3x+7y−17)=0
⇒x(3λ+1)+y(7λ+2)−(17λ+5)=0⋯(1)
Slope of the line
m1=−(3λ+17λ+2)
Now, slope of the other given line
3x+4y−10=0m2=−34
As the lines are perpendicular, so
m1m2=−1⇒−(3λ+17λ+2)(−34)=−1⇒9λ+3=−28λ−8⇒λ=−1137
Putting the value of λ in equation (1), we get
x(−3337+1)+y(−7737+2)−(−18737+5)=0
∴4x−3y+2=0
Alternate solution:
Given line 3x+4y−10=0
Perpendicular line to this
4x−3y+μ=0⋯(2)
Now finding the point of intersection of
x+2y−5=0 and 3x+7y−17=0
We get
x=5−2y⇒3(5−2y)+7y−17=0⇒y=2⇒x=1
Using equation (2), we get
4−6+μ=0⇒μ=2
So, the equation of the required line is
4x−3y+2=0