The line `(x-2)/3=(y+1)/2=(z-1)/-1` intersects the curve `x y=c^(2),z=0` if `c` is equal to a. `+-1` b. `+-1/3` c. `+-sqrt(5)` d. none of these

Options

A. `+-1`
B. `+-1//3`
C. `+-sqrt(5)`
D. none of these

Answer

C

Explanation

At the point on the line where it intersects the given curve, we have `z=0` so that
`(x-2)/3=(y+1)/2=(0-1)/(-1)`
`implies(x-2)/3=1` and `(y+1)/2=1impliesx=5` and `y=1.`
Putting these values of `x` and `y` in `xy=c^(2)`, we get `c^(2)=5impliesc=+-sqrt(5)`.