If the lines ax + y + 1 = 0, x + by + 1 = 0 a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) If the given lines are concurrent,the determinant of their coefficients must be zero:$\begin{vmatrix} a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c \end{vma

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) If the given lines are concurrent,the determinant of their coefficients must be zero:$\begin{vmatrix} a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c \end{vmatrix} = 0$Applying column operations $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$:$\begin{vmatrix} a & 1-a & 1-a \ 1 & b-1 & 0 \ 1 & 0 & c-1 \end{vmatrix} = 0$Expanding the determinant:$a(b-1)(c-1) - (1-a)(c-1) - (1-a)(b-1) = 0$Dividing the entire equation by $(1-a)(1-b)(1-c)$ (since $a, b, c 
eq 1$):$\frac{a(b-1)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(b-1)}{(1-a)(1-b)(1-c)} = 0$$\frac{-a}{(1-a)} + \frac{1}{(1-b)} + \frac{1}{(1-c)} = 0$Adding $1$ to both sides:$1 - \frac{a}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$$\frac{1-a+a}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$

If the lines $ax + y + 1 = 0, x + by + 1 = 0$ and $x + y + c = 0$ ($a, b, c$ being distinct and different from $1$) are concurrent,then $\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = $

Similar Questions

The straight lines $x + 2y - 9 = 0$,$3x + 5y - 5 = 0$ and $ax + by - 1 = 0$ are concurrent,if the straight line $35x - 22y + 1 = 0$ passes through the point

The value of $k$ such that the lines $2x - 3y + k = 0$,$3x - 4y - 13 = 0$,and $8x - 11y - 33 = 0$ are concurrent,is

If $3a + 5b + 6c = 0$,then the family of lines $ax + by + c = 0$ passes through the fixed point:

The lines $2x + y - 1 = 0$,$ax + 3y - 3 = 0$,and $3x + 2y - 2 = 0$ are concurrent for

Let $\alpha, \beta, \gamma$ be three non-zero real constants and $a, b, c$ be three arbitrary real numbers which satisfy $\alpha a + \beta b + \gamma c = 0$. Then the point of concurrence of the family of lines $ax + by + c = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module