Question

p/a+q/b+r/c=1and a/p+b/q+c/r=0 so prove that p^2/a^2+q^2/b^2r^2/c^2=1​

Answer 1

Given : p/a+q/b+r/c=1  $\frac{p}{a} +\frac{q}{b} + \frac{r}{c} = 1$   and a/p+b/q+c/r=0    $\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0$  

To find : Prove that p^2/a^2+q^2/b^2r^2/c^2=1​    $\frac{p^2}{a^2} +\frac{q^2}{b^2}+ \frac{r^2}{c^2} = 1$

Solution:  

$\frac{p}{a} +\frac{q}{b} + \frac{r}{c} = 1$

Squaring both sides

$\implies ( \frac{p}{a} +\frac{q}{b} + \frac{r}{c} )^2= 1^2$

Using (x + y + z)² = x² + y² + z²  + 2(xy + yz  + zx)  

$\implies ( \frac{p}{a})^2 +(\frac{q}{b})^2 + (\frac{r}{c} )^2 + 2(\frac{pq}{ab} +\frac{qr}{bc} + \frac{pr}{ac})= 1$

Taking pqr/abc  common from bracket

$\implies ( \frac{p}{a})^2 +(\frac{q}{b})^2 + (\frac{r}{c} )^2 + 2(\frac{pqr}{abc} ) (\frac{c}{r} +\frac{a}{p} + \frac{b}{q})= 1$

$\implies ( \frac{p}{a})^2 +(\frac{q}{b})^2 + (\frac{r}{c} )^2 + 2(\frac{pqr}{abc} ) (\frac{a}{p} + \frac{b}{q} + \frac{c}{r} )= 1$

Substitute given value of  $\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0$  

$\implies ( \frac{p}{a})^2 +(\frac{q}{b})^2 + (\frac{r}{c} )^2 + 2(\frac{pqr}{abc} ) (0 )= 1$

$\implies ( \frac{p}{a})^2 +(\frac{q}{b})^2 + (\frac{r}{c} )^2 + 0 = 1$

$\implies ( \frac{p}{a})^2 +(\frac{q}{b})^2 + (\frac{r}{c} )^2 = 1$

$\implies ( \frac{p^2}{a^2}) +(\frac{q^2}{b^2}) + (\frac{r^2}{c^2} ) = 1$

$\implies \frac{p^2}{a^2} +\frac{q^2}{b^2}+ \frac{r^2}{c^2} = 1$

QED

Hence Proved

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