Question

find the distance of a point (3,5)from the line2x+3y=14 measured prallel to the line x-2y=1.

Answer 1

use parametric form of a straight line that is
x-a/cos=y-b/sin=+-r
a,b are 3,5
tan =1/2 so sin is 1/√3 and cos is 2/√3

Answer 2

Parallel line to (x-2y = 1) , which passes through pune (3,5) is ;
x-2y+k =0
3-(2×5)+k = 0
3-10 = -k
k = 7
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Hence the equation of line is ;
x-2y+7 = 0

To find the point at which this two lines inherent ;

x-2y+7 = 0 -----(multiply by 2)

_ 2x-4y = -14
2x+3y = 14
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0-7y = -28
y = 4

x = 2y-7
= (2×4)-7
= 1

Hence the point in the line 2x+3y-14=0 is
P (1,4)

Distance between the point and the line
= Distance between points (3,5) & (1,4)
= root{(3-1)^2 + (5-4)^2}
= root (4+1)
= √5
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