Reduce the equation x/a +y/b=1 in perpendicular form and hence prove :1/p^2=1/a^2+1/b^2

Answer 1

The given equation is the intercept form. The x- and y- intercepts are (a,0) and (0,b).

So, what do you know about the altitude of a right triangle, drawn to the hypotenuse?

Answer 2

here,
(bx+ay)/ab=1
or, (bx+ay0=ab

Dividing both sides by +-√(x-coefficient)^2+(y-coefficient)^2
=+-√(b)^2+(a)^2
=+-√(a^2)+(b^2)
∴ (bx)/{+-√(a^2)+(b^2)} +(ay)/{+-√(a^2)+(b^2)} =(ab)/{+-√(a^2)+(b^2)}

Comparing with x-cos alpha + y-sin alpha = perpendicular
perperndicular=(ab)/{+-√(a^2)+(b^2)}

Squaring on both sides,

P^2=[(ab)/{+-√(a^2)+(b^2)}]^2
or, P^2=(a^2×b^2)/{(a^2) + (b^2)}
or, {(a^2) + (b^2)}/(a^2×b^2)=1/p^2
or, (a^2)/(a^2×b^2)+(b^2)/(a^2×b^2)=1/p^2
or, 1/p^2 = 1/(b^2) + 1/(a^2)

HENCE, 1/p^2 = 1/(b^2) + 1/(a^2) PROVED#

Answer 3

here,
(bx+ay)/ab=1
or, (bx+ay)=ab

Dividing both sides by +-√(x-coefficient)^2+(y-coefficient)^2
=+-√(b)^2+(a)^2
=+-√(a^2)+(b^2)
∴ (bx)/{+-√(a^2)+(b^2)} +(ay)/{+-√(a^2)+(b^2)} =(ab)/{+-√(a^2)+(b^2)}

Comparing with x-cos alpha + y-sin alpha = perpendicular
perperndicular=(ab)/{+-√(a^2)+(b^2)}

Squaring on both sides,

P^2=[(ab)/{+-√(a^2)+(b^2)}]^2
or, P^2=(a^2×b^2)/{(a^2) + (b^2)}
or, {(a^2) + (b^2)}/(a^2×b^2)=1/p^2
or, (a^2)/(a^2×b^2)+(b^2)/(a^2×b^2)=1/p^2
or, 1/p^2 = 1/(b^2) + 1/(a^2)

HENCE, 1/p^2 = 1/(b^2) + 1/(a^2) PROVED#