1.

from \( A \).(i) Find the distance of the point \( (3,5) \) from the line \( 2 x+3 y=14 \) measured parallel to the line \( x-2 y=1 \)

Answer»

Let the equation of the line parallel to x−2y=1 is x−2y+λ=0
Since, it passes through (3,5)
⇒3−10+λ=0
⇒λ=7
Therefore, the line is x−2y+7=0.
The point of intersection of x−2y+7=0 and 2x+3y−14=0 is (1,4).
The distance between (3,5) and (1,4)

\(=\sqrt{3-1)^2+(5-4)^2}=\sqrt{4+1}=\sqrt5.\)



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