Examples on Concurrency of Three Lines Set 2

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Example – 11

Show that the equation of any line passing through the intersection point P of two given lines whose equations are L1 and L2, 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 L1 and L2 vanish.

Thus, \({L_1} + \lambda {L_2} = 0\) is the equation of an arbitrary straight line that passes through the intersection point P of L1 and L2. (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 L1 and L2 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 L1 and L2 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 :

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

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