Question

If $a,b,c$ are the $p^{th},q^{th}$ and $r^{th}$ terms of an $H.P$, then the lines $bcx+py+1=0,cax+qy+1=0$ and $abx+ry+1=0$,

• A. are concurrent
• B. form a triangle
• C. are parallel
• D. mutually perpendicular lines

Answer 1

The correct option is A are concurrent

Given $a,b,c$ are $p^{th},q^{th},r^{th}$ terms of H.P. respectively.

$\Rightarrow \frac{1}{a},\frac{1}{b},\frac{1}{c}$ are $p^{th},q^{th},r^{th}$ terms of A.P. respectively.

Let $A$ be the first term and d be the common difference

$T_p=A+(p-1)d$

$\frac{1}{a}=A+(p-1)d$ .....(1)

$T_q=A+(q-1)d$

$\frac{1}{b}=A+(q-1)d$ .....(2)

$T_r=A+(r-1)d$

$\frac{1}{c}=A+(r-1)d$ .....(3)

Subtracting (2) from (1), we get

$\frac{1}{a}-\frac{1}{b}=(p-q)d$

$\Rightarrow ab(p-q)=\frac{b-a}{d}$ .....(4)

Subtracting (3) from (2), we get

$\frac{1}{b}-\frac{1}{c}=(q-r)d$

$\Rightarrow bc(q-r)=\frac{c-b}{d}$ .....(5)

Subtracting (3) from (1), we get

$\frac{1}{a}-\frac{1}{c}=(p-r)d$

$\Rightarrow ac(p-r)=\frac{c-a}{d}$ .....(6)

Now, consider $\left|\begin{array}{ccc}bc& p& 1\\ ca& q& 1\\ ab& r& 1\end{array}\right|$

(The above matrix is the matrix representation of the given equations)

Expanding along first column, we get

$=bc(q-r)-ca(p-r)+ab(p-q)$

$=\frac{c-b}{d}-\frac{c-a}{d}+\frac{b-a}{d}$

$=0$

Hence, the given lines are concurrent.

Hence, option A.