If p is the length of the perpendicular from the origin to the line `(x)/(a) + (y)/(b) = 1, "then prove that " (1)/(p^(2)) = (1)/(a^(2)) + (1)/(b^(2))`

Options

A. `p^(2)=a^(2)+b^(2)`
B. `(1)/(p^(2))=(1)/(a^(2))+(1)/(b^(2))`
C. `p=a+b`
D. `(1)/(p)=(1)/(a)+(1)/(b)`

Answer

B

Explanation

The given line is `bx+ay-ab=0" '(i)`
p=Lenth of the perpendicular from the origin to (i)
`therefore p=(|bxx0+axx0-ab|)/(sqrt(b^(2)+a^(2)))=(ab)/(sqrt(a^(2)+b^(2)))`
`implies p^(2) =(a^(2)b^(2))/(a^(2)+b^(2))implies (1)/(p^(2))=(1)/(a^(2))+(1)/(b^(2))`