Question -
Answer -
Given:
The equation isparallel to 3x − 4y + 5 = 0 and pass through (2, 3)
The equation of theline parallel to 3x − 4y + 5 = 0 is
3x – 4y+ λ = 0,
Where, λ isa constant.
It passes through (2,3).
Substitute the valuesin above equation, we get
3 (2) – 4 (3) + λ = 0
6 – 12 + λ =0
λ = 6
Now, substitute thevalue of λ = 6 in 3x – 4y + λ = 0, we get
3x − 4y + 6
∴ The required line is3x − 4y + 6 = 0.