Answer :

The given line is 2x + 3y = 14 = L1

Line parallel to x –2y = 1 which passes through (3,5) is 3–2(5) = k

K = –7

So, equation of line becomes x –2y = –7 = L2

The intersection point of L1 and L2

2x + 3y = 14

x –2y = –7

Multiplying L2 by 2 and the n L1 – L2, we get,

x = 1 and y = 4

So, distance between (1,4) and (3,5)

Using distance formula

So, distance of the point (3, 5) from the line 2x + 3y–14 = 0 measured parallel to the line x–2y = 1 is 5 units

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