If `(a,b)` and `(c,d)` are two points on the whose equation is `y=mx+k`, then the distance between `(a,b)` and `(c,d)` in terms of `a,c` and `m` is

Options

A. `|a+c|sqrt(1+m^(2))`
B. `|a-c|sqrt(1+m^(2))`
C. `(|a-c|)/(sqrt(1+m^(2))`
D. `|a-c|(a+m^(2))`

Answer

B

Explanation

Let `m=tantheta`
then equation of line through `(a,b)` is
`(x-a)/(costheta)=(y-b)/(sintheta)=r`
if `(c,d)` lies on it at `r` distance then
`impliesc-a=rcosthetaimpliesr-(c-a)sectheta`
`impliesr=|c-a|sqrt(1+m^(2))`
Alter: Since `(a,b)` and `(c,d)` are on the same line `y=mx+k`, they satisfy the same equation. Therefore `b=ma+kd=mc+k`.
Now the distance between `(a,b)` and `(c,d)` is
`sqrt((a-c)^(2)+(b-d)^(2))`
From the first to equations we obtain `(b-d)=m(a-c)`, so that
`sqrt((a-c)^(2)+(b-d)^(2))=sqrt((a-c)^(2)+m^(2)(a-c)^(2))`
`=|a-c|sqrt(1+m^(2))`
Note we are using the fact that `=sqrt(x^(2))=|x|` for all real `x`.