The line L given by `x/5+y/b=1` passes through the point (13, 32). The line K is parallel to L and has the equation `x/c+y/3=1` Then the distance between L and K is

Options

A. `23/sqrt15`
B. `sqrt17`
C. `17/sqrt15`
D. `23/sqrt17`

Answer

D

Explanation

Since, line L passes through (13,32)
` 13/5 + 32/b =1`
` Rightarrow 32/b = 1- 13/5 Rightarrow 32/b = -8/5 `
` Rightarrow b=- (32xx5)/8 = -20`
` Rightarroow L : x/5-y/20 =1`
Given ` K = x/c + y/3 =1` is parallel to L = 0
The line K must have equation `x/5 -y/20 =a`
` x /(5a) -y/(20a) =1`
On comparing with ` x/c + y/3 =1` , we get
` -20 a =3 and c=5a`
` Rightarrow a = -3/20 and c= -15/20`
Distance between lines is
`|(a-1)/(sqrt(1/25+1/400))| = |(-3/20 -1)/(sqrt(17/400))| = 23/sqrt17`