Lines in planes - class-XII

A system of homogeneous linear equations has non-trivial solutions if the determinant of the coefficient matrix is zero. The determinant of [[1, 2, 3], [2, 3, 4], [7, 13, 19]] is 1(57-52) - 2(38-28) + 3(26-21) = 5 - 20 + 15 = 0. Since the determinant is zero, non-trivial solutions exist.

For the system to have non-trivial solutions, the determinant must be zero. The determinant of [[2, -3, 1], [1, 2, -3], [4, -1, -2]] is 2(-4-3) - (-3)(-2+12) + 1(-1-8) = 2(-7) + 3(10) - 9 = -14 + 30 - 9 = 7. Since the determinant is not zero, the system has only the trivial solution (0,0,0).

The given equations passes through same point.So, they are concurrent lines

$\Rightarrow \left| \begin{matrix} { \lambda  }^{ 2 } & -1 & -1 \ 1 & -{ \lambda  }^{ 2 } & 1 \ 1 & 1 & -{ \lambda  }^{ 2 } \end{matrix} \right| =0$

$\Rightarrow { \lambda  }^{ 6 }-3{ \lambda  }^{ 2 }-2=0$

Since $7x,36y,12z$ are divisible by $k$

Here coefficient matrix,  $\Delta = \begin{vmatrix} 2 & 1 & -1 \ a & 3 & -3 \ 3 & 2 & -2 \end{vmatrix}$
Using $C _2 \to C _2+C _3$
$\Delta = \begin{vmatrix} 2 & 0 & -1 \ a & 0 & 3 \ 3 & 0 & -2 \end{vmatrix}=0$
Clearly $\Delta=0$, Hence given lines are concurrent for all values of $a$

Given lines $\displaystyle y-x=5,3x+4y=1$ and $\displaystyle y=mx+3$
For concurrency,
$\begin{vmatrix} -1 & 1 & 5 \ 3 & 4 & 1 \ -m & 1 & 3 \end{vmatrix}=0$
$\Rightarrow -5+19m=0$
$\Rightarrow \displaystyle m =\frac{5}{19}$

These lines are concurrent when area formed from these lines is zero
$\begin{vmatrix} p & q & r \ q & r & p \ r & p & q \end{vmatrix}=0$
Applying ${ R } _{ 1 }\rightarrow { R } _{ 1 }+{ R } _{ 2 }+{ R } _{ 3 }$
$\Rightarrow \begin{vmatrix} p+q+r & p+q+r & p+q+r \ q & r & p \ r & p & q \end{vmatrix}=0\ \Rightarrow \left( p+q+r \right) \begin{vmatrix} 1 & 1 & 1 \ q & r & p \ r & p & q \end{vmatrix}=0$
Again applying ${ C } _{ 2 }\rightarrow { C } _{ 2 }-{ C } _{ 1 },{ C } _{ 3 }\rightarrow { C } _{ 3 }-{ C } _{ 1 }$
$\Rightarrow \left( p+q+r \right) \begin{vmatrix} 1 & 0 & 0 \ q & r-q & p-q \ r & p-r & q-r \end{vmatrix}\ \Rightarrow \left( p+q+r \right) \left( \left( r-q \right) \left( q-r \right) -\left( p-r \right) \left( p-q \right)  \right) =0$<br>$\Rightarrow p^3+q^3+r^3-3pqr =0 $
Hence, options A, B and C are correct.

$2x-ay+1 =0,\ 3x-by+1 =0,\ 4x-cy+1 =0$

The given points are collinear if 

$x+2y-9=0$ ---(1)

$3x+5y-5=0$ ---(2)

$ax+by-1=0$ ---(3)

Solving (1) and (2) simultaneously we get 

$y=22, x=-35$

Now, equation (1), (2) and (3) will be concurrent, that is they will pass through one point if $y=22, x=-35$ satisfy the third equation $ax+by-1=0$.

For the lines to be concurrent, the determinant of the coefficients must be zero: |a, b, c; b, c, a; c, a, b| = 0. This expands to -(a^3 + b^3 + c^3 - 3abc) = 0, implying a+b+c=0 or a=b=c. Given a!=b!=c, we must have a+b+c=0. The exponent is (a^2/bc + b^2/ca + c^2/ab) = (a^3+b^3+c^3)/abc. Since a^3+b^3+c^3 = 3abc when a+b+c=0, the exponent is 3. Thus, 2^3 = 8.

First we need to know if the above equations are linearly dependent or no,

$p _{1}x+q _{1}y=1,\ p _{2}x+q _{2}y =1\ p _{3}x+q _{3}y=1$
Given lines are concurrent
$\Rightarrow \begin{vmatrix} p _{1} & q _{1} & 1 \ p _{2} & q _{2}& 1 \ p _{3} & q _{3} & 1 \end{vmatrix}=0$

$\Rightarrow p _{1}(q _{2}-q _{3})-q _{1}(p _{2}-p _{3})+(p _{2}q _{3}-p _{3}q _{2})=0$

$\Rightarrow (p _{1}q _{2}-p _{2}q _{1})+(p _{2}q _{3}-p _{3}q _{2})+(p _{3}q _{1}-p _{1}q _{3})=0$
The left hand side of the above equation is also equal to twice the area of a triangle with coordinates $(p _1, q _1),; (p _2, q _2),; (p _3,q _3)$
Since it is equal to zero, $(p _{1},q _{1}),(p _{2},q _{2}),(p _{3},q _{3})$ are collinear.

$\Delta =\begin{vmatrix} x+1 & x+2 & x+a \ x+2 & x+3 & x+b \ x+3 & x+4 & x+c \end{vmatrix}=0$