Question

if the lines ax + y + 1 = 0 , x + by + 1 =0 and x + y + c =0 where a,b and c are distinct real numbers different from 1 are constant concurrent , then the value of 1/ 1-a + 1/ 1-b + 1/ 1-c equals what?

Answer 1

Check the Attachment

Answer 2

The value of $\frac{1}{1-a}+\frac{1}{1-b} +\frac{1}{1-c}$ is 1.

• Given equations are

                  ax + y + 1 = 0   --------------(1)

                  x + by + 1 = 0   --------------(2)

                   x + y + c = 0    --------------(3)

• Coefficient matrix of (1), (2), (3) is

                 $A=\left[\begin{array}{ccc}a&1&1\\1&b&1\\1&1&c\end{array}\right]$

• (1), (2), (3) are concurrent

                 ⇒ |A| = 0

                 ⇒ det$\left[\begin{array}{ccc}a&1&1\\1&b&1\\1&1&c\end{array}\right] =0$

                 $R_{2}$→$R_{2}-R_{1},R_{3}$→$R_{3}-R_{1}$

                 ⇒ det $\left[\begin{array}{ccc}a&1&1\\1-a&b-1&0\\1-a&0&c-1\end{array}\right] =0$

                 ⇒ a(b-1)(c-1)-(1-a)(c-1)+(-(1-a)(b-1))=0

                 ⇒ a(1-b)(1-c)+(1-a)(1-c)+(1-a)(1-b)=0

       Dividing by (1-a)(1-b)(1-c) on both sides, we get

                 ⇒ $\frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0$

                 ⇒ $\frac{1+a-1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0$

                 ⇒ $\frac{1}{1-a}-\frac{1-a}{1-a}+\frac{1}{1-b} +\frac{1}{1-c}=0$

                 ⇒ $\frac{1}{1-a}-1+\frac{1}{1-b} +\frac{1}{1-c}=0$

                 ⇒ $\frac{1}{1-a}+\frac{1}{1-b} +\frac{1}{1-c}=1$

   

Note: Please use determinant symbol where I mentioned det along with matrix.