Given:
The equation is perpendicular to x – 3y + 5 = 0 and passes through (3,-2)
The equation of the line perpendicular to x − 3y + 5 = 0 is
3x + y + λ = 0,
Where, λ is a constant.
It passes through (3, − 2).
Substitute the values in above equation, we get
3 (3) + (-2) + λ = 0
9 – 2 + λ = 0
λ = – 7
Now, substitute the value of λ = − 7 in 3x + y + λ = 0, we get
3x + y – 7 = 0
The required line is 3x + y – 7 = 0.
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