Question

If p is the length of the perpendicular from the origin to the line `x/a+y/b=1`,prove that `1/(p²)=1/(a²)+1/(b²)`?

Answer 1

See below.

Explanation:

Considering the triangle

`BOA` with

`B = (0,b)`
`O=(0,0)`
`A = (a,0)`

we have

`bar(OB)^2-p^2=n^2`
`bar(OA)^2-p^2 = m^2`
`p^2=n m`

then

`(bar(OB)^2-p^2)(bar(OA)^2-p^2)=n^2m^2=p^4`

developping

`bar(OB)^2bar(OA)^2-bar(OB)^2 p^2-bar(OA)^2 p^2=0` or

`p^2 = (bar(OB)^2bar(OA)^2)/(bar(OA)^2+bar(OB)^2)` or finally

`1/p^2 = 1/bar(OA)^2+1/bar(OB)^2 = 1/a^2+1/b^2`