# If p is the length of perpendicular from the origin on the line xa+yb = 1 and a2, p2, b2 are in A.P, then show that a4 + b4 = 0. - Mathematics

Sum

If p is the length of perpendicular from the origin on the line x/a + y/b = 1 and a2, p2, b2 are in A.P, then show that a4 + b4 = 0.

#### Solution

Given equation is x/a + y/b = 1

p = |(0/a + 0/b - 1)/sqrt(1/a^2 + 1/b^2)|

Squaring both sides, we have

p2 = 1/(1/a^2 + 1/b^2)

⇒ 1/a^2 + 1/b^2 = 1/p^2   ......(i)

Since a2, p2, b2 are in A.P.

∴ 2p2 = a2 + b2

⇒ p2 = (a^2 + b^2)/2

⇒ 1/p^2 = 2/(a^2 + b^2)

Putting the value of 1/p^2 is equation (i) we get,

1/a^2 + 1/b^2 = 2/(a^2  + b^2)

⇒ (a^2 + b^2)/(a^2b^2) = 2/(a^2 + b^2)

⇒ (a2 + b2)2 = 2a2b2

⇒ a4 + b4 + 2a2b2 = 2a2b2

⇒ a4 + b4 = 0.

Hence proved.

Concept: Slope of a Line
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#### APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 10 Straight Lines
Exercise | Q 21 | Page 180
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