# Examples on Concurrency of Three Lines Set 2

**Example – 11**

Show that the equation of any line passing through the intersection point *P* of two given lines whose equations are *L*_{1} and *L*_{2}, can be expressed as \({L_1} + \lambda \,{L_2} = 0,\,\,{\rm{where}}\,\,\lambda \) is a real parameter.

**Solution: ** Let \({L_1} \equiv {a_1}x + {b_1}y + {c_1} = 0\) and \({L_2} \equiv {a_2}x + {b_2}y + {c_2} = 0\)

Consider the equation \({L_1} + \lambda \,{L_2} = 0 \qquad \qquad \qquad \qquad \qquad \qquad \dots\rm{(1)}\)

\[\begin{align} \Rightarrow & \qquad {a_1}x + {b_1}y + {c_1} + \lambda \left( {{a_2}x + {b_2}y + {c_2}} \right) = 0\\ \Rightarrow & \qquad \left( {{a_1} + \lambda {a_2}} \right)x + \left( {{b_1} + \lambda {b_2}} \right)y + \left( {{c_1} + \lambda {c_2}} \right) = 0 \end{align}\]

This is definitely the equation of a straight line because it is of the form \(ax + by + c = 0.\) Also, notice in addition that the intersection point *P* will satisfy this equation, because if we substitute the co-ordinates of the intersection point *P* in (1), both *L*_{1} and *L*_{2} vanish.

Thus, \({L_1} + \lambda {L_2} = 0\) is the equation of an arbitrary straight line that passes through the intersection point *P* of *L*_{1} and *L*_{2}. (As we vary \(\lambda ,\) the slope of this line will vary but it will **always** pass through *P*).

This result is very beneficial in certain cases. We’ll see such cases in some subsequent examples

**Example – 12**

Find the equations to the straight lines passing through

**(a)** (3, 2) and the point of intersection of \(2x + 3y = 1 \text{ and } 3x - 4y = 6\)

**(b)** Origin and the point of intersection of \(\begin{align}\frac{x}{a} + \frac{y}{b} = 1{\text{ and }} \frac{x}{b} + \frac{y}{a} = 1\end{align}.\)

**Solution:****(a)** The equations of the two given lines in standard form are :

\[\begin{align}&{L_1}\,\,\,\, \equiv \,\,\,2x + 3y - 1 = 0\\&{L_2}\,\,\,\, \equiv \,\,\,3x - 4y - 6 = 0\end{align}\]

Any line passing through the intersection point of *L*_{1} and *L*_{2} is

\[\begin{align}&{L_1} + \lambda {L_2} = 0\\\Rightarrow \qquad &(2x + 3y - 1) + \lambda (3x - 4y - 6) = 0\\\Rightarrow \qquad &(2 + 3\lambda )x + (3 - 4\lambda )y - (1 + 6\lambda ) = 0 \qquad \qquad\qquad \dots \rm{(1)} \end{align} \]

We want this line to pass through (3, 2). Therefore (3, 2) must satisfy the equation of this line, i.e.

\[\begin{align}&(2 + 3\lambda )3 + (3 - 4\lambda )2 - (1 + 6\lambda ) = 0\\\Rightarrow \quad &- 5\lambda + 11 = 0\\\Rightarrow \quad &\lambda = \frac{{11}}{5}\end{align}\]

We substitute \(\begin{align}\lambda = \frac{{11}}{5}\end{align}\) in (1) to get the required equation:

\[\begin{align}&(2 + 2 \cdot \frac{{11}}{5})x + (3 - 4 \cdot \frac{{11}}{5})y - (1 + 6 \cdot \frac{{11}}{5}) = 0\\\Rightarrow \qquad &43x - 29y - 71 = 0\end{align}\]

**(b)** We follow the same procedure as in part (a)

\[\begin{align}{L_1}\,\,\,\,:\,\,\,bx + ay - ab = 0\\{L_2}\,\,\,\,:\,\,\,ax + by - ab = 0\end{align}\]

The equation of any line passing through the intersection point of *L*_{1} and *L*_{2} is

\[\begin{align}&{L_1} + \lambda {L_2} = 0\\\Rightarrow \qquad &(b + \lambda a)x + (a + \lambda b)y - ab(1 + \lambda ) = 0 \qquad \qquad \dots \rm{(2)}\end{align}\]

Since this line must pass through (0, 0), we substitute (0, 0) into (2) to get

\[\begin{align}&ab(1 + \lambda ) = 0\\\Rightarrow \qquad &\lambda = - 1\end{align}\]

We substitute \(\lambda = - 1\) into (2) to get the required equation :

\[\begin{align}&(b - a)x + (a - b)y = 0\\\Rightarrow \qquad &x - y = 0\end{align}\]