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  • The base BC of a triangle ABC contains the points <img alt="\mathrm{P\left(p_1, q_1\right) and Q\left(p_2, q_2\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BP%5Cleft%28p

The base BC of a triangle ABC contains the points \mathrm{P\left(p_1, q_1\right) and Q\left(p_2, q_2\right)}  and the equation of sides A B and A C are \mathrm{p_1 x+q_1 y=1 } and \mathrm{q_2 x+p_2 y=1} respectively. The equation of A Q is
and  \mathrm{\left(k p_2 q_2-1\right)\left(p_1 x+q_1 y-1\right)=\left(p_1 p_2+q_1 q_2-1\right)\left(q_2 x+p_2 y-1\right)}  where k=.
 

Option: 1

2


Option: 2

-2


Option: 3

1


Option: 4

-1


Answers (1)

best_answer


Since both A P and A Q pass through A the intersection of the two given lines, their equation is

\mathrm{\left(p_1 x+q_1 y-1\right)+\lambda\left(q_2 x+p_2 y-1\right)=0(1) }

In order to find the equation of A Q we find the value of \lambda by the fact that equation (1) passes through \mathrm{Q\left(p_2, q_2\right)}

\begin{array}{cl} & \mathrm{\left(p_1 p_2+q_1 q_2-1\right)+\lambda\left(2 p_2 q_2-1\right)=0 }\\ \Rightarrow \quad \lambda_2= & \mathrm{-\frac{\left(p_i p_2+q_j q_2-1\right)}{\left(2 p_2 q_2-l\right)}} \end{array}

Therefore equation of A Q is \mathrm{\left(2 p_2 q_2-1\right)\left(p_1 x+q_1 y-1\right)=\left(p_1 p_2+q_1 q_2-1\right)\left(q_2 x+p_2 y-\right. 1)}

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Deependra Verma

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