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, show that the points $(p_1,q_1),(p_2,q_2)$ and $p_3,q_3)$ are coolinear.

Answer 1

If the lines are concurrent, then the lines have common point of intersection.

The given line are

$p_{1}x+q_{1}y=1...\left(1\right)$

$p_{2}x+q_{2}y=1...\left(2\right)$

$p_{3}x+q_{3}y=1...\left(3\right)$

Solving (1) and (2)

$x=\frac{1-q_{1}y}{p_1}$

$=p_2\left(\frac{1-q_1,y}{p_1}\right)+q_{2}y=1$

$p_2=p_2{q}_{1}y+p_1{q}_{2}y=p_1$

$y=\frac{p_1{p}_2}{p_1{p}_2-p_2{q}_1}$

$\Rightarrow x=\frac{1-q_1\left(\frac{p_1-p_2}{p_1{p}_2-p_2{q}_1}\right)}{p_1}$

Putting x, y in (3)

$p_3\left[\right(p_1{q}_2-p_2{q}_1)-q_1{p}_1-q_1{p}_2][p_1{q}_2-p_2{q}_1]+q_3{p}_1(p_1-p_2)=1$

$(p_1{p}_3{q}_2-p_2{p}_3{q}_1-p_1{p}_3{q}_1+p_2{p}_3{q}_1)+q_3{p}_1^2-q_3{p}_1{p}_2=1$

$(p_1{p}_3{q}_2-p_1{p}_3{q}_1)(p_1{q}_2-p_2{q}_1)+q_3{p}_1^2-q_3{p}_1{p}_2=1$

$p_1^2{p}_3{q}_2^2-p_1{p}_2{p}_3{q}_1{q}_2-p_1^2{p}_3{q}_2{q}_1+p_1{p}_2{p}_3{q}_1^2+q_3{p}_1^2-q_3{p}_1{p}_2=1\left(1\right)$

Also if $\left(p_1{q}_1\right)\left(p_2{p}_2\right)\left(p_3{p}_3\right)$ are collinear

Then,

$p_1(q_2-q_3)+p_2(q_3-q_1)+p_3(q_1-q_3)=0$

From (1)

$p_1[p_1{p}_3{q}_2^2-p_2{p}_3{q}_1{q}_2-p_1{p}_3{q}_1{q}_2+p_2{p}_3{q}_1^2+q_3{p}_1-q_3{p}_2]=1$

$p_1\left[p_3{q}_2\right(p_1{p}_2-p_2{q}_1)-p_3{q}_1(p_1{p}_2-p_2{q}_1)+q_3(p_1-p_2\left)\right]=1$

Hence, the points are collinear.