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  • Explain Solution R. D. Sharma Class 12 Chapter deteminants Exercise 5.2 Question 50 maths Textbook Solution.

Explain Solution R. D. Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 50 maths Textbook Solution.

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Answer: 2

Hint Use determinant formula

Given: $$ \left|\begin{array}{lll} p & b & c \\ q & q & c \\ a & b & r \end{array}\right|=0 \text { Find value of } \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} \& p \neq a, q \neq b, r \neq c

Solution:

$$ \begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} p & b & c \\ q & q & c \\ a & b & r \end{array}\right| \\ &\text { Apply } \mathrm{R}_{1} \rightarrow R_{1}-R_{3} \& R_{2} \rightarrow R_{2}-R_{3} \\ &\left|\begin{array}{ccc} p-a & 0 & c-p \\ 0 & q-b & c-p \\ a & 0 & r \end{array}\right|=0 \end{aligned}

$$ \begin{aligned} &\text { Expand from } \mathrm{R}_{1}\\ &=p-a\left|\begin{array}{cc} q-b & c-p \\ b & r \end{array}\right|+(c-r)\left|\begin{array}{cc} 0 & q-b \\ a & b \end{array}\right|=0\\ &=(p-a)[r(q-b)-b(c-r)+(c-r)(-a(q-b))]=0\\ &=\frac{r(q-b)(p-a)}{(p-a)(q-b)(r-c)}-\frac{b(p-a)(c-r)}{(p-a)(q-b)(r-c)}-\frac{a(c-r)(q-b)}{(p-a)(q-b)(r-c)}=0 \end{aligned}

$$ \begin{aligned} &=\frac{r}{r-c}+\frac{b}{q-b}+\frac{a}{p-a}=0 \\ &=\frac{r}{r-c}+\frac{b+q-q}{q-b}+\frac{a+p-p}{p-a}=0 \\ &=\frac{r}{r-c}+\frac{b-q}{q-b}+\frac{q}{q-b}+\frac{a-p}{p-a}-\frac{p}{p-a}=0 \\ &=\frac{r}{r-c}-1+\frac{q}{q-b}-1-\frac{p}{p-a}=0 \\ &=\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=2 \end{aligned}

 

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