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

Answer:

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?

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

Please refer to the adjoining figure

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.