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

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

Question

(B) For three lines to be concurrent,the determinant of their coefficients must be zero:$\left| \begin{array}{ccc} a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) For three lines to be concurrent,the determinant of their coefficients must be zero:$\left| \begin{array}{ccc} a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c \end{array} ight| = 0$Applying column operations $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$:$\left| \begin{array}{ccc} a & 1-a & 1-a \ 1 & b-1 & 0 \ 1 & 0 & c-1 \end{array} ight| = 0$Expanding along the first row:$a(b-1)(c-1) - (1-a)(c-1) - (1-a)(b-1) = 0$Divide the entire equation by $(1-a)(1-b)(1-c)$:$\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$$-\frac{a}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0$Since $-\frac{a}{1-a} = \frac{1-a-1}{1-a} = \frac{1}{1-a} - 1$,we substitute this into the equation:$(\frac{1}{1-a} - 1) + \frac{1}{1-b} + \frac{1}{1-c} = 0$$\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 $x + y + c = 0$ (where $a, b$ and $c$ are distinct and different from $1$) are concurrent,then the value of $\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = $

Similar Questions

The straight lines $4ax + 3by + c = 0$,where $a + b + c = 0$,will be concurrent at the point:

Three lines $3x - y = 2$,$5x + ay = 3$,and $2x + y = 3$ are concurrent. Find the value of $a$.

If $x+2y-3=0$,$3x+4y-7=0$,$2x+3y-4=0$,and $4x+5y-6=0$ are the equations of four lines,then

If three lines whose equations are $y=m_{1}x+c_{1}$,$y=m_{2}x+c_{2}$,and $y=m_{3}x+c_{3}$ are concurrent,then show that $m_{1}(c_{2}-c_{3})+m_{2}(c_{3}-c_{1})+m_{3}(c_{1}-c_{2})=0$.

For the family of lines given by the equation $(2 + k)x + (1 + k)y = 5 + 7k$,which of the following statements is true for different values of $k$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module