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  • The base of a triangle <img alt="\mathrm{AB C}" src="https://entrancecorner.oncodecogs.com/gif.

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

\mathrm{and \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\: \mathrm{k}=.}

Option: 1

2


Option: 2

-2


Option: 3

1


Option: 4

-1


Answers (1)

Since both \mathrm{A P \: and\: A Q} pass through \mathrm{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 \mathrm{ A Q } we find the value of \mathrm{ \lambda }  by the fact that equation (1) passes through \mathrm{ Q\left(p_2, q_2\right) }

\mathrm{ \left(p_1 p_2+q_1 q_2-1\right)+\lambda\left(2 p_2 q_2-1\right)=0 }

\mathrm{ \Rightarrow \lambda=-\frac{\left(p_1 p_2+q_1 q_2-I\right)}{\left(2 p_2 q_2-I\right)} }

Therefore equation of \mathrm{ A Q \: is \: \: \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)}

Hence option 1 is correct.

Posted by

Kshitij

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