Prove that the line `x=ay+b, z=cy+d and x=a'y+b', z=dy+d'` are perpendicular if `aa'+cc'+1=0`

Options

A. `aa'+cc'=1`
B. `aa'+"cc"'=-1`
C. `ab+cd=a'b'+c'd'`
D. `aa'+"bb"'="cc"'+dd'`

Answer

B

Explanation

The equations of the given linear are not in symmetrical form. We first put them in symmetrical form.
Equations of first line are `x=ay+b,z=cy+d`. These equations can be written as
`(x-b)/a=y,(z-d)/c=yimplies(x-b)/a=(y-0)/1=(z-d)/c`……………i
Similarly `x=a', y+b',z=c', y+d'` can be written as
`(x-b')/(a')=(y-0)/1=(z-d')/(c')`.............ii
If lines i and ii are perpendicular then
`aa'+1+"cc"'=0impliesaa'+"cc"'=-1`