Question

If the lines $p_{1}x+q_{1}y=1,p_{2}x+q_{2}y=1$ and $p_{3}x+q_{3}y=1$ be concurrent, then the points $\left(p_1,q_1\right),\left(p_2,q_2\right)$ and $\left(p_3,q_3\right)$,

• A. are collinear
• B. form an equilateral triangle
• C. form a scalene triangle
• D. form a right angled triangle

Answer 1

Answer: (A)

$p_{1}x+q_{1}y=1,p_{2}x+q_{2}y=1p_{3}x+q_{3}y=1$
Given lines are concurrent
$\Rightarrow \left|\begin{array}{ccc}{p}_1& q_1& 1\\ p_2& q_2& 1\\ p_3& q_3& 1\end{array}\right|=0$
$\Rightarrow p_1\left(q_2-q_3\right)-q_1\left(p_2-p_3\right)+\left(p_2{q}_3-p_3{q}_2\right)=0$
$\Rightarrow \left(p_1{q}_2-p_2{q}_1\right)+\left(p_2{q}_3-p_3{q}_2\right)+\left(p_3{q}_1-p_1{q}_3\right)=0$
The left hand side of the above equation is also equal to twice the area of a triangle with coordinates
$\left(p_1,q_1\right),\left(p_2,q_2\right),\left(p_3,q_3\right)$
Since it is equal to zero, $\left(p_1,q_1\right),\left(p_2,q_2\right),\left(p_3,q_3\right)$ are collinear.