Question -
Answer -
Given:
The equation isperpendicular to x – 3y + 5 = 0 and passes through (3,-2)
The equation of theline perpendicular to x − 3y + 5 = 0 is
3x + y + λ =0,
Where, λ isa constant.
It passes through(3, − 2).
Substitute the valuesin above equation, we get
3 (3) + (-2) +λ = 0
9 – 2 + λ =0
λ = – 7
Now, substitute thevalue of λ = − 7 in 3x + y + λ = 0, we get
3x + y – 7 = 0
∴ The required line is3x + y – 7 = 0.