SOLUTION: if p is the length if the perpendicular from the origin to the line x/a + y/b = 1, prove that 1/p² = 1/a² + 1/b². I assume I'll have to use the perpendicular distance formula so

Algebra ->  Test -> SOLUTION: if p is the length if the perpendicular from the origin to the line x/a + y/b = 1, prove that 1/p² = 1/a² + 1/b². I assume I'll have to use the perpendicular distance formula so      Log On


   

Question 698546: if p is the length if the perpendicular from the origin to the line x/a + y/b = 1, prove that 1/p² = 1/a² + 1/b².
I assume I'll have to use the perpendicular distance formula somehow, but I don't really know what to do ...
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
The formula for the perpendicular distance from (x1,y1)
to the line whose equation is Ax+By+C=0 is

d = abs%28%28Ax%5B1%5D%2BBy%5B1%5D%2BC%29%2Fsqrt%28A%5E2%2BB%5E2%29%29

The line has the equation

x%2Fa + y%2Fb = 1  

[Don't get capital A and B confused with little letters a and b]

Clear of fractions by multiplying by LCD = ab:

     bx + ay = ab

Get 0 on the right

bx + ay - ab = 0

For (x1,y1) = (0,0)

Substitute in

d = abs%28%28Ax%5B1%5D%2BBy%5B1%5D%2BC%29%2Fsqrt%28A%5E2%2BB%5E2%29%29

with d = p, A = b, B = a, C = -ab, (x1,y1) = (0,0)

p = abs%28%28b%2A%280%29%2Ba%2A%280%29-ba%29%2Fsqrt%28b%5E2%2Ba%5E2%29%29

p = abs%28-ba%2Fsqrt%28b%5E2%2Ba%5E2%29%29

Square both sides

p² = b%5E2a%5E2%2F%28b%5E2%2Ba%5E2%29

Take the reciprocal of both sides:

1%2Fp%5E2 = %28b%5E2%2Ba%5E2%29%2F%28b%5E2a%5E2%29

Make two fractions on the right:

1%2Fp%5E2 = b%5E2%2F%28b%5E2a%5E2%29%2Ba%5E2%2F%28b%5E2a%5E2%29

1%2Fp%5E2 = 

1%2Fp%5E2 = 1%2Fa%5E2%2B1%2Fb%5E2

Edwin