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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

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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.

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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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