Question

20. The lines (p-q) x + (q-r) y +(r-p) = 0
(q-r)x+(r-p) y+(p-q)=0, (r-p)x+(p-9)y+(q-r)=0

Can someone help me with this question
I know the answer ,I want to know the method.


I will mark you as the brqinliest

Answer 1

Answer:

Concept used:

If the lines

$ax+by+c=0,\:lx+my+n=0,\:px+qy+r=0$ are concurrent, then

$\left|\begin{array}{ccc}a&b&c\\l &m&n\\p&q&r\end{array}\right|=0$

Given lines are

(p-q)x+(q-r)y+(r-p) = 0

(q-r)x+(r-p) y+(p-q)=0,

(r-p)x+(p-9)y+(q-r)=0

$\left|\begin{array}{ccc}p-q&q-r&r-p\\q-r &r-p&p-q\\r-p&p-q&q-r\end{array}\right|$

$=\left|\begin{array}{ccc}0&0&0\\q-r &r-p&p-q\\r-p&p-q&q-r\end{array}\right|\:R_1\implies\:R_1+R_2+R_3$

$=0$

$\therefore\:\text{The given lines are concurrent}$

Answer 2

here is ur answer...