if p denotes the length of perpendicular line drawn from origin to a line x÷a+y÷b=1 then show that 1÷a^2+1÷b^2=I ÷p^2
The equation is,
x/a + y/b = 1 rearranging,
y = (-b/a)x + b
this is a straight line of slope m = -b/a and y-intercept c = b
if two lines, y = mx + c and y2 = m2.x + c2 are mutually perpindicular then m.m2 = -1
Since m = -b/a, the the slope of the perpindicular line is m2 = a/b
Since the line y2 = m2.x + c2 passes through the origin then c2 = 0. Then,
y2 = (a/b)x
The intersection of y and y2
when y = y2, then
-(b/a)x + b = (a/b)x
x(a/b + b/a) = b
x(a^2 + b^2)/ab = b
x = ab^2/(a^2 + b^2)
Since y = (a/b).x, then
y = a^2.b/(a^2 + b^2)
The distance p, from the origin to the point of intersection, is the square root of the sum of the squares of the x- and y-coordinates, i.e.
p^2 = x^2 + y^2
p^2 = a^2b^4/(a^2 + b^2)^2 + a^4b^2/(a^2 + b^2)^2
p^2 = (a^2b^4 + a^4b^2)/(a^2 + b^2)^2
p^2 = a^2b^2(b^2 + a^2)/(a^2 + b^2)^2
p^2 = a^2b^2/(a^2 + b^2)
1/p^2 = (a^2 + b^2)/(a^2b^2)
1/p^2 = 1/b^2 + 1/a^2