Class 11: The Straight Line

$\displaystyle \textbf{Question 1: }\text{Prove that the following sets of three lines are concurrent:}$
$\displaystyle \text{i) }15x-18y+1=0,\quad 12x+10y-3=0\text{ and }6x+66y-11=0$
$\displaystyle \text{ii) }3x-5y-11=0,\quad 5x+3y-7=0\text{ and }x+2y=0$
$\displaystyle \text{iii) }\frac{x}{a}+\frac{y}{b}=1,\quad \frac{x}{b}+\frac{y}{a}=1\text{ and }y=x,\quad a,b\neq0$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{i) The given lines are}$
$\displaystyle 15x-18y+1=0,\quad 12x+10y-3=0\text{ and }6x+66y-11=0.$
$\displaystyle \text{Three lines }a_1x+b_1y+c_1=0,\ a_2x+b_2y+c_2=0\text{ and }a_3x+b_3y+c_3=0$
$\displaystyle \text{are concurrent if }\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}=0.$
$\displaystyle \begin{vmatrix}15&-18&1\\12&10&-3\\6&66&-11\end{vmatrix}$
$\displaystyle =15\begin{vmatrix}10&-3\\66&-11\end{vmatrix}+18\begin{vmatrix}12&-3\\6&-11\end{vmatrix}+\begin{vmatrix}12&10\\6&66\end{vmatrix}$
$\displaystyle =15(-110+198)+18(-132+18)+(792-60)$
$\displaystyle =1320-2052+732=0.$
$\displaystyle \therefore \text{The given three lines are concurrent.}$
$\displaystyle \\$

$\displaystyle \text{ii) The given lines are}$
$\displaystyle 3x-5y-11=0,\quad 5x+3y-7=0\text{ and }x+2y=0.$
$\displaystyle \begin{vmatrix}3&-5&-11\\5&3&-7\\1&2&0\end{vmatrix}$
$\displaystyle =3\begin{vmatrix}3&-7\\2&0\end{vmatrix}+5\begin{vmatrix}5&-7\\1&0\end{vmatrix}-11\begin{vmatrix}5&3\\1&2\end{vmatrix}$
$\displaystyle =3(14)+5(7)-11(10-3)$
$\displaystyle =42+35-77=0.$
$\displaystyle \therefore \text{The given three lines are concurrent.}$
$\displaystyle \\$

$\displaystyle \text{iii) The given equations are}$
$\displaystyle \frac{x}{a}+\frac{y}{b}=1,\quad \frac{x}{b}+\frac{y}{a}=1\text{ and }y=x.$
$\displaystyle \text{Multiplying the first two equations by }ab,\text{ we get}$
$\displaystyle bx+ay-ab=0,\quad ax+by-ab=0\text{ and }x-y=0.$
$\displaystyle \begin{vmatrix}b&a&-ab\\a&b&-ab\\1&-1&0\end{vmatrix}$
$\displaystyle =b\begin{vmatrix}b&-ab\\-1&0\end{vmatrix}-a\begin{vmatrix}a&-ab\\1&0\end{vmatrix}-ab\begin{vmatrix}a&b\\1&-1\end{vmatrix}$
$\displaystyle =-ab^2-a^2b-ab(-a-b)$
$\displaystyle =-ab^2-a^2b+a^2b+ab^2=0.$
$\displaystyle \therefore \text{The given three lines are concurrent.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{For what value of }\lambda,\text{ are the three lines}$
$\displaystyle 2x-5y+3=0,\quad 5x-9y+\lambda=0\text{ and }x-2y+1=0\text{ concurrent?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Given lines are }2x-5y+3=0,\quad 5x-9y+\lambda=0\text{ and }x-2y+1=0.$
$\displaystyle \text{Since the lines are concurrent,}$
$\displaystyle \begin{vmatrix}2&-5&3\\5&-9&\lambda\\1&-2&1\end{vmatrix}=0.$
$\displaystyle \Rightarrow 2(-9+2\lambda)+5(5-\lambda)+3(-10+9)=0.$
$\displaystyle \Rightarrow -18+4\lambda+25-5\lambda-3=0.$
$\displaystyle \Rightarrow 4-\lambda=0.$
$\displaystyle \therefore \lambda=4.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the condition that the straight lines}$
$\displaystyle y=m_1x+c_1,\quad y=m_2x+c_2\text{ and }y=m_3x+c_3\text{ may meet in a point.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{The given lines can be written as}$
$\displaystyle m_1x-y+c_1=0,\quad m_2x-y+c_2=0\text{ and }m_3x-y+c_3=0.$
$\displaystyle \text{For the three lines to be concurrent,}$
$\displaystyle \begin{vmatrix}m_1&-1&c_1\\m_2&-1&c_2\\m_3&-1&c_3\end{vmatrix}=0.$
$\displaystyle \Rightarrow m_1(c_2-c_3)+(m_2c_3-m_3c_2)+c_1(m_3-m_2)=0.$
$\displaystyle \Rightarrow m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0.$
$\displaystyle \therefore \text{The required condition is}$
$\displaystyle m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{If the lines }p_1x+q_1y=1,\quad p_2x+q_2y=1\text{ and}$
$\displaystyle p_3x+q_3y=1\text{ are concurrent, show that the points }(p_1,q_1),\ (p_2,q_2)$
$\displaystyle \text{and }(p_3,q_3)\text{ are collinear.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{The given lines can be written as}$
$\displaystyle p_1x+q_1y-1=0,\quad p_2x+q_2y-1=0\text{ and }p_3x+q_3y-1=0.$
$\displaystyle \text{Since the three lines are concurrent,}$
$\displaystyle \begin{vmatrix}p_1&q_1&-1\\p_2&q_2&-1\\p_3&q_3&-1\end{vmatrix}=0.$
$\displaystyle \text{Taking }-1\text{ common from the third column,}$
$\displaystyle -\begin{vmatrix}p_1&q_1&1\\p_2&q_2&1\\p_3&q_3&1\end{vmatrix}=0.$
$\displaystyle \Rightarrow \begin{vmatrix}p_1&q_1&1\\p_2&q_2&1\\p_3&q_3&1\end{vmatrix}=0.$
$\displaystyle \text{This is the condition for the points }(p_1,q_1),\ (p_2,q_2)\text{ and }(p_3,q_3)$
$\displaystyle \text{to be collinear.}$
$\displaystyle \therefore \text{The points }(p_1,q_1),\ (p_2,q_2)\text{ and }(p_3,q_3)\text{ are collinear.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Show that the straight lines }L_1=(b+c)x+ay+1=0,$
$\displaystyle L_2=(c+a)x+by+1=0\text{ and }L_3=(a+b)x+cy+1=0$
$\displaystyle \text{are concurrent, where }a+b+c\neq0.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{The given lines are}$
$\displaystyle L_1=(b+c)x+ay+1=0,$
$\displaystyle L_2=(c+a)x+by+1=0,$
$\displaystyle L_3=(a+b)x+cy+1=0.$
$\displaystyle \text{Let }x=y=-\frac{1}{a+b+c}.$
$\displaystyle \text{For }L_1,\quad (b+c)x+ay+1$
$\displaystyle =-\frac{b+c}{a+b+c}-\frac{a}{a+b+c}+1$
$\displaystyle =-\frac{a+b+c}{a+b+c}+1=0.$
$\displaystyle \text{Similarly,}$
$\displaystyle (c+a)x+by+1=0\quad\text{and}\quad(a+b)x+cy+1=0.$
$\displaystyle \text{Therefore, all three lines pass through the common point}$
$\displaystyle \left(-\frac{1}{a+b+c},-\frac{1}{a+b+c}\right).$
$\displaystyle \therefore \text{The given straight lines are concurrent.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{If the three lines }ax+a^2y+1=0,\quad bx+b^2y+1=0\text{ and}$
$\displaystyle cx+c^2y+1=0\text{ are concurrent, show that at least two of the three constants}$
$\displaystyle a,\ b,\ c\text{ are equal.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{The given lines are}$
$\displaystyle ax+a^2y+1=0,\quad bx+b^2y+1=0\text{ and }cx+c^2y+1=0.$
$\displaystyle \text{Since the three lines are concurrent,}$
$\displaystyle \begin{vmatrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{vmatrix}=0.$
$\displaystyle \text{Applying }R_1\rightarrow R_1-R_2\text{ and }R_2\rightarrow R_2-R_3,$
$\displaystyle \begin{vmatrix}a-b&a^2-b^2&0\\b-c&b^2-c^2&0\\c&c^2&1\end{vmatrix}=0.$
$\displaystyle \Rightarrow (a-b)(b-c)\begin{vmatrix}1&a+b&0\\1&b+c&0\\c&c^2&1\end{vmatrix}=0.$
$\displaystyle \Rightarrow (a-b)(b-c)\begin{vmatrix}1&a+b\\1&b+c\end{vmatrix}=0.$
$\displaystyle \Rightarrow (a-b)(b-c)\{(b+c)-(a+b)\}=0.$
$\displaystyle \Rightarrow (a-b)(b-c)(c-a)=0.$
$\displaystyle \therefore a=b\quad\text{or}\quad b=c\quad\text{or}\quad c=a.$
$\displaystyle \therefore \text{At least two of the constants }a,\ b,\ c\text{ are equal.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{If }a,\ b,\ c\text{ are in A.P. and }3a-2b\neq0,\text{ prove that the}$
$\displaystyle \text{straight lines }ax+2y+1=0,\quad bx+3y+1=0\text{ and }cx+4y+1=0$
$\displaystyle \text{are concurrent.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Since }a,\ b,\ c\text{ are in A.P.,}$
$\displaystyle 2b=a+c.$
$\displaystyle \text{The determinant of the coefficients of the three lines is}$
$\displaystyle \begin{vmatrix}a&2&1\\b&3&1\\c&4&1\end{vmatrix}.$
$\displaystyle \text{Applying }R_1\rightarrow R_1-R_2\text{ and }R_2\rightarrow R_2-R_3,$
$\displaystyle \begin{vmatrix}a-b&-1&0\\b-c&-1&0\\c&4&1\end{vmatrix}$
$\displaystyle =\begin{vmatrix}a-b&-1\\b-c&-1\end{vmatrix}$
$\displaystyle =-(a-b)+(b-c)$
$\displaystyle =2b-a-c.$
$\displaystyle \text{Since }2b=a+c,$
$\displaystyle 2b-a-c=0.$
$\displaystyle \therefore \begin{vmatrix}a&2&1\\b&3&1\\c&4&1\end{vmatrix}=0.$
$\displaystyle \text{Also, }3a-2b\neq0,\text{ so the first two lines are not parallel.}$
$\displaystyle \therefore \text{The three straight lines are concurrent.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Show that the perpendicular bisectors of the sides of a triangle are concurrent.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let }ABC\text{ be a non-degenerate triangle.}$
$\displaystyle \text{Let the perpendicular bisectors of }AB\text{ and }BC\text{ intersect at }O.$
$\displaystyle \text{Since }O\text{ lies on the perpendicular bisector of }AB,$
$\displaystyle OA=OB.$
$\displaystyle \text{Since }O\text{ lies on the perpendicular bisector of }BC,$
$\displaystyle OB=OC.$
$\displaystyle \therefore OA=OC.$
$\displaystyle \text{Hence, }O\text{ is equidistant from }A\text{ and }C.$
$\displaystyle \text{Therefore, }O\text{ lies on the perpendicular bisector of }AC.$
$\displaystyle \text{Thus, the perpendicular bisectors of }AB,\ BC\text{ and }CA$
$\displaystyle \text{pass through the same point }O.$
$\displaystyle \therefore \text{The perpendicular bisectors of the sides of a triangle are concurrent.}$
$\displaystyle \\$