Question 1: Prove that the following sets of three lines are concurrent:

i) $15 x - 18 y + 1 = 0, \ \ 12 x + 10 y - 3 = 0$ and $6 x + 66 y - 11 = 0$

ii) $3 x -5 y -11 =0, \ \ 5 x + 3 y -7 =0$ and $x +2y =0$

iii) $\frac{x}{a}$ $+$ $\frac{y}{b}$ $=1 \ \$ $\frac{x}{b}$ $+$ $\frac{y}{a}$ $=1$ and $y = x$

i)       Given lines are $15 x - 18 y + 1 = 0$$12 x + 10 y - 3 = 0$  and $6 x + 66 y - 11 = 0$

We have:

$\begin{vmatrix} 15 & -18 & 1 \\ 12 & 10 & -3 \\ 6 & 66 & -11 \end{vmatrix}$

$= 15 ( - 110 + 198) + 18 ( -132 +18) + 1 ( 792 - 60)$

$= 1320 - 2052 + 732 = 0$

Hence, the given lines are concurrent.

ii)      Given equations: $3 x -5 y -11 =0, \ \ 5 x + 3 y -7 =0$ and $x +2y =0$

We have:

$\begin{vmatrix} 3 & -5 & 11 \\ 5 & 3 & -7 \\ 1 & 2 & 0 \end{vmatrix}$

$= 3 \times 14 + 5 \times 7 - 11 \times 7$

$= 42+35-77 = 0$

Hence, the given lines are concurrent.

iii)    Given equations: $\frac{x}{a}$ $+$ $\frac{y}{b}$ $=1 \ \$ $\frac{x}{b}$ $+$ $\frac{y}{a}$ $=1$ and $y = x$

or the given equations are $bx + ay = ab \ \ ax + by = ab$ and $y = x$

We have:

$\begin{vmatrix} b & a & -ab \\ a & b & -ab \\ 1 & -1 & 0 \end{vmatrix}$

$= -b \times ab -a \times ab - ab ( -a-b)$

$= -ab^2 - a^2b + a^b + ab^2 = 0$

Hence, the given lines are concurrent.

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Question 2: For what value of  $\lambda$ , are the three lines $2 x -5 y + 3 = 0, \ \ 5 x - 9 y + \lambda = 0$ and $x - 2 y + 1 = 0$ concurrent?

Given lines are $2 x -5 y + 3 = 0, \ \ 5 x - 9 y + \lambda = 0$ and $x - 2 y + 1 = 0$

Since the lines are concurrent,

$\begin{vmatrix} 2 & -5 & 3 \\ 5 & -9 & \lambda \\ 1 & -2 & 1 \end{vmatrix} = 0$

$\Rightarrow 2( -9 + 2 \lambda) + 5 ( 5-\lambda) + 3 ( -10 + 9 ) = 0$

$\Rightarrow -18 + 4 \lambda + 25 - 5 \lambda - 3 = 0$

$\Rightarrow \lambda = 4$

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Question 3: Find the conditions that the straight lines $y = m_1 x + c_1, \ \ y = m_2x + c_2$ and $y = m_3 x + c_3$ may meet in a point.

Given lines are $y = m_1 x + c_1, \ \ y = m_2x + c_2$ and $y = m_3 x + c_3$

For the lines to meet at a point or for the lines to be concurrent,

$\begin{vmatrix} m_1 & -1 & c_1 \\ m_2 & -1 & c_2 \\ m_3 & -1 & c_3 \end{vmatrix} = 0$

$\Rightarrow m_1( -c_3+c_2) + 1 ( m_2c_3 - m_3 c_2) + c_1( -m_2 + m_3) = 0$

$\Rightarrow m_1( c_2-c_3) + m_2( c_3 - c_1) + m_3 ( c_1 - c_2)=0$

Hence, the required condition is $m_1( c_2-c_3) + m_2( c_3 - c_1) + m_3 ( c_1 - c_2)=0$  for the given lines to meet at a point.

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Question 4: If the lines $p_1x+q_1y= 1, \ \ p_2x+q_2y= 1$ and $p_3x+q_3y= 1$ be concurrent! show that the points $( p_1, q_1) , (p_2, q_2)$ and $(p_3, q_3)$ are collinear.

The given lines are $p_1x+q_1y= 1, \ \ p_2x+q_2y= 1$ and $p_3x+q_3y= 1$

If these lines are concurrent, then

$\begin{vmatrix} p_1 & q_1 & -1 \\ p_2 & q_2 & -1 \\ p_3 & q_3 & -1 \end{vmatrix} = 0$

$\Rightarrow (-1) \begin{vmatrix} p_1 & q_1 & 1 \\ p_2 & q_2 & 1 \\ p_3 & q_3 & 1 \end{vmatrix} = 0$

$\Rightarrow \begin{vmatrix} p_1 & q_1 & 1 \\ p_2 & q_2 & 1 \\ p_3 & q_3 & 1 \end{vmatrix} = 0$

Which is the condition of collinearity of three points $( p_1, q_1) , (p_2, q_2)$ and $(p_3, q_3)$

Hence, if the given lines are concurrent, then the points are collinear.

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Question 5: Show that the straight lines $L_1 = (b+c)x + ay + 1 =0,$ $L_2 = (c+a)x + by + 1 =0$ and  $L_3 = (a+b)x + cy + 1 =0$ are concurrent.

Given lines are $L_1 = (b+c)x + ay + 1 =0,$ $L_2 = (c+a)x + by + 1 =0$ and  $L_3 = (a+b)x + cy + 1 =0$

If these lines are concurrent, then the determinant should be equal to $0$

We have $\begin{vmatrix} b+c & a & 1 \\ c+a & b & 1\\ a+b & c & 1 \end{vmatrix}$

Applying transformation $C_1 \rightarrow C_1+C_2$

$= \begin{vmatrix} a+b+c & a & 1 \\ c+a+b & b & 1\\ a+b +c & c & 1 \end{vmatrix}$

$= (a+b+c) \begin{vmatrix} 1 & a & 1 \\ 1 & b & 1\\ 1 & c & 1 \end{vmatrix}$

Applying transformation $C_1 \rightarrow C_1 - C_2$

$= (a+b+c) \begin{vmatrix} 0 & a & 1 \\ 0 & b & 1\\ 0 & c & 1 \end{vmatrix}$

$= 0$

Therefore the given lines are concurrent.

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Question 6: If the three lines $ax+a^2y +1 = 0 , bx+b^2 y + 1 = 0$ and $cx + c^2 y + 1 = 0$ are concurrent, show that at least two of three constants $a, b, c$ are equal.

Given lines are $ax+a^2y +1 = 0 , bx+b^2 y + 1 = 0$ and $cx + c^2 y + 1 = 0$

Since the given lines are concurrent,

$\begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1\\ c & c^2 & 1 \end{vmatrix} = 0$

Applying transformation $R_1 \rightarrow R_1 - R_2, R_2 \rightarrow R_2 - R_3$

$\Rightarrow \begin{vmatrix} a-b & a^2-b^2 & 0 \\ b-c & b^2-c^2 & 0 \\ c & c^2 & 1 \end{vmatrix} = 0$

$\Rightarrow ( a-b)(b-c) \begin{vmatrix} 1 & a+b & 0 \\ 1 & b+c & 0 \\ c & c^2 & 1 \end{vmatrix} = 0$

$\Rightarrow (a-b)(b-c) (c-a) = 0$

$\therefore a=b \text{ or } b=c \text{ or } c=a$

Therefore at least two of three constants $a, b, c$ are equal

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Question 7: If $a, b, c$ are in A.P., prove that the straight lines $ax+2y + 1 = 0 , \ \ bx+3y +1 = 0$  and $cx + 4y +1 = 0$  are concurrent.

Given $a, b, c$ are in A.P.    $\Rightarrow 2b = a+c$     … … … … … i)

Given lines are $ax+2y + 1 = 0 , \ \ bx+3y +1 = 0$  and $cx + 4y +1 = 0$

For the lines to be concurrent, the determinant should be $0$. Therefore,

$\begin{vmatrix} a & 2 & 1 \\ b & 3 & 1\\ c & 4 & 1 \end{vmatrix}$

Applying transformation $R_1 \rightarrow R_1 - R_2, \ \ \ R_2 \rightarrow R_2 - R_3$

$\Rightarrow \begin{vmatrix} a-b & -1 & 0 \\ b-c & -1 & 0 \\ c & 4 & 1 \end{vmatrix}$

$\Rightarrow -a+b+b-c$

$\Rightarrow 2b - ( a+c)$

Substituting i) we get

$\Rightarrow 2b - 2b = 0$

Hence the lines are concurrent.

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Question 8: Show that the perpendicular bisectors of the Sides of a triangle are concurrent.

Let the $\triangle ABC$ have the vertices $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$

Let $D, E$ and $F$ be the midpoints of line $BC, CA$ and $AB$ respectively.

Therefore the coordinates of $D, E$ and $F$ are

$D \Big($ $\frac{x_2+x_3}{2}$ $,$ $\frac{y_2+y_3}{2}$ $\Big), D \Big($ $\frac{x_1+x_3}{2}$ $,$ $\frac{y_1+y_3}{2}$ $\Big)$ and $D \Big($ $\frac{x_1+x_2}{2}$ $,$ $\frac{y_1+y_2}{2}$ $\Big)$

Slope of $BC =$ $\frac{y_3-y_2}{x_3-x_2}$

Therefore the slope of $AD = -$ $\frac{x_3-x_2}{y_3-y_2}$

Therefore the equation of $AD$:

$y -$ $\frac{y_2+y_3}{2}$ $= -$ $\frac{x_3-x_2}{y_3-y_2}$ $\Big( x -$ $\frac{x_2+x_3}{2}$ $\Big)$

$\Rightarrow 2y ( y_3-y_2) - ( {y_3}^2 - {y_2}^2) = - 2x ( x_3 - x_2) + ( {x_3}^2 - {x_2}^2 )$

$\Rightarrow 2x ( x_3 - x_2) + 2y ( y_3-y_2) - ( {x_3}^2 - {x_2}^2 ) - ( {y_3}^2 - {y_2}^2) = 0$     … … … … … i)

Similarly, the  equations of BE and CF are

$2x ( x_1 - x_3) + 2y ( y_1-y_3) - ( {x_1}^2 - {x_3}^2 ) - ( {y_1}^2 - {y_3}^2) = 0$     … … … … … ii)

$2x ( x_2 - x_1) + 2y ( y_2-y_1) - ( {x_2}^2 - {x_1}^2 ) - ( {y_2}^2 - {y_1}^2) = 0$     … … … … … iii)

For lines i), ii) and iii) to be concurrent, the determinant should be $0$. Therefore,

$\begin{vmatrix} 2( x_3 - x_2) & 2( y_3-y_2) & - ( {x_3}^2 - {x_2}^2 ) - ( {y_3}^2 - {y_2}^2) \\ 2( x_1 - x_3) & 2( y_1-y_3) & - ( {x_1}^2 - {x_3}^2 ) - ( {y_1}^2 - {y_3}^2) \\ 2( x_2 - x_1) & 2( y_2-y_1) & - ( {x_2}^2 - {x_1}^2 ) - ( {y_2}^2 - {y_1}^2) \end{vmatrix}$

Applying transformation $R_1 \rightarrow R_1+R_2+R_3$

$= \begin{vmatrix} 0 & 0 & 0 \\ 2( x_1 - x_3) & 2( y_1-y_3) & - ( {x_1}^2 - {x_3}^2 ) - ( {y_1}^2 - {y_3}^2) \\ 2( x_2 - x_1) & 2( y_2-y_1) & - ( {x_2}^2 - {x_1}^2 ) - ( {y_2}^2 - {y_1}^2) \end{vmatrix}$

$= 0$

Hence the three perpendicular bisectors are concurrent.