Straight Lines
11.0 Family of straight lines
11.0 Family of straight lines
The equation of a family of straight lines passing through the point of intersection of the lines $${L_1} \equiv {a_1}x + {b_1}y + {c_1} = 0$$ and $${L_2} \equiv {a_2}x + {b_2}y + {c_2} = 0$$ is given by $${L_1} + \lambda {L_2} = 0$$ i.e., $$({a_1}x + {b_1}y + {c_1}) + \lambda ({a_2}x + {b_2}y + {c_2}) = 0$$ where $\lambda $ is an arbitrary real number.
Question 7. Find the equation of straight line which passes through the point $(2,-3)$ and the point of intersection of the lines $x+y+4=0$ and $3x-y-8=0$.
Solution: Any line passes through the intersection of two lines $x+y+4=0$ and $3x-y-8=0$ can be written in the form $$(x + y + 4) + \lambda (3x - y - 8) = 0$$ which passes through $(2,-3)$. Therefore, $$\begin{equation} \begin{aligned} (2 - 3 + 4) + \lambda (6 + 3 - 8) = 0 \\ 3 + \lambda = 0 \\ \lambda = - 3 \\\end{aligned} \end{equation} $$
Putting the value of $\lambda $, we get $$\begin{equation} \begin{aligned} (x + y + 4) - 3(3x - y - 8) = 0 \\ 2x - y - 7 = 0 \\\end{aligned} \end{equation} $$ which is the equation of required line.
