The value of $P_{1}^{- 2} + P_{2}^{- 2} + P_{3}^{- 2}$ is equal to

\frac{s - a}{Δ}

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\frac{a + b + c}{Δ}

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\frac{a^{2} + b^{2} + c^{2}}{4 Δ^{2}}

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\frac{a^{2} + b^{2} + c^{2}}{Δ^{2}}

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Solution

The correct option is D $\frac{a^{2} + b^{2} + c^{2}}{4 Δ^{2}}$
$P_{1}, P_{2}, P_{3}$ is perpendicular on $B C, C A, A B$
And $B C = a, C A = b, A B = C$
Therefore area of triangle is $\frac{1}{2}$ XperpendicularXbase
$△ = \frac{1}{2} P_{1} . a = \frac{1}{2} P_{2} . b = \frac{1}{2} P_{3} . c P_{1} = \frac{2 △}{a}, P_{2} = \frac{2 △}{b}, P_{3} = \frac{2 △}{c} \frac{1}{{P_{1}}^{2}} + \frac{1}{{P_{2}}^{2}} + \frac{1}{{P_{3}}^{2}} = \frac{a^{2} + b^{2} + c^{2}}{4 △^{2}}$

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The value of $\frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \frac{1}{r_{3}^{2}} + \frac{1}{r^{2}}$ is

Δ, \frac{a^{2} + b^{2} + c^{2}}{R}

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b^{2} = a^{2} + c^{2}

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

Δ = 27

a^{2} + b^{2} + c^{2} = 2

Δ

A B C

A, B, C

a, b, c

(a^{2} - b^{2} - c^{2}) tan A + (a^{2} - b^{2} + c^{2})

tan B

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