Solution:
Comparing each of them with r ⃗ . ( a i ^ + b j ^ ) + c = 0 \vec r.(a\hat i+b\hat j)+c=0 r . ( a i ^ + b j ^ ) + c = 0
we get,
a 1 = 4 , b 1 = 3 , c 1 = − 1 a 2 = − 2 , b 2 = 1 , c 2 = − 1 a 3 = 1 , b 3 = 2 , c 3 = − 1 a_1=4, b_1=3,c_1=-1
\\ a_2=-2, b_2=1,c_2=-1
\\ a_3=1,b_3=2,c_3=-1 a 1 = 4 , b 1 = 3 , c 1 = − 1 a 2 = − 2 , b 2 = 1 , c 2 = − 1 a 3 = 1 , b 3 = 2 , c 3 = − 1
For these lines to be concurrent, they must satisfy following:
∣ a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 ∣ = 0 \begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3 \\
\end{vmatrix} =0 ∣ ∣ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ∣ ∣ = 0
Take LHS
= ∣ a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 ∣ =\begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3 \\
\end{vmatrix} = ∣ ∣ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ∣ ∣
Putting above values
= ∣ 4 3 − 1 − 2 1 − 1 1 2 − 1 ∣ =\begin{vmatrix}
4 & 3 & -1 \\
-2 & 1 & -1 \\
1 & 2 & -1 \\
\end{vmatrix} = ∣ ∣ 4 − 2 1 3 1 2 − 1 − 1 − 1 ∣ ∣
= 4 ∣ 1 − 1 2 − 1 ∣ − 3 ∣ − 2 − 1 1 − 1 ∣ + ( − 1 ) ∣ − 2 1 1 2 ∣ =4\begin{vmatrix}
1 & -1 \\
2 & -1 \\
\end{vmatrix}-3\begin{vmatrix}
-2 & -1 \\
1 & -1 \\
\end{vmatrix}+(-1)\begin{vmatrix}
-2 & 1 \\
1 & 2 \\
\end{vmatrix} = 4 ∣ ∣ 1 2 − 1 − 1 ∣ ∣ − 3 ∣ ∣ − 2 1 − 1 − 1 ∣ ∣ + ( − 1 ) ∣ ∣ − 2 1 1 2 ∣ ∣
= 4 ( − 1 + 2 ) − 3 ( 2 + 1 ) − 1 ( − 4 − 1 ) = 4 − 9 + 5 = 9 − 9 = 0 =4(-1+2)-3(2+1)-1(-4-1)
\\ =4-9+5
\\ =9-9
\\ =0 = 4 ( − 1 + 2 ) − 3 ( 2 + 1 ) − 1 ( − 4 − 1 ) = 4 − 9 + 5 = 9 − 9 = 0
= = = RHS
Thus, above lines are concurrent.
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