Q. 75.0( 1 Vote )

The distance of t

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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