Unit-I: Sets and Functions
Unit-II: Algebra
Unit-III: Coordinate Geometry
Unit-IV: Calculus
Unit-V: Mathematical Reasoning
Unit-VI: Statistics and Probability

Condition of coincidence of lines

  L_1 :a_1x+b_1y+c_1=0

 L_2 :a_2x+b_2y+c_2=0

  L_1   and  L_2  coincidence lines if  \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}

But in parallel lines, the condition satisfy  \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}

Point of Intersection:

If two disticnt lines  L_1   and  L_2 , represented by the equations 

 L_1 : a_1x+b_1y+c_1=0

 L_2 : a_2x+b_2y+c_2=0 intersect at   P(h.k)

then   h=\dfrac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}, \quad k=\dfrac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1} \text { where } a_1b_2-a_2b_1 \neq0


 L_1 : a_1x+b_1y+c_1=0

 L_2 : a_2x+b_2y+c_2=0    then

(i) if  \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}   then  L_1    and  L_2   intersect at one point and it have unique solution.

(ii) If  \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} , then  L_1    and  L_2    never intesect, where  L_1    and  L_2   are parallel lines and having no solution.

(iii) If  \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} =\dfrac{c_1}{c_2} , then  L_1    and  L_2    are coincidenc linea and having infinitel many solutions.

Examples for Corresponding Conditions:

(i)  L_1 : 2x+3y-9=0, \quad L_2 :5x-3y+2=0

(ii)  L_1 : 4x+8=0, \quad L_2 :2x+4y+3=0

(iii)  L_1 : 2x-3y+7=0, \quad L_2 :6x-9y+21=0

Family of lines through the point of intersection of two lines:

Let  L_1 : a_1x+b_1y+c_1=0   -(i)

 L_2 :a_2x+b_2y+c_2=0 -(ii)

Now consider the equation  (a_1x+b_1y+c_1)+\lambda (a_2x+b_2y+c_2) =0   -(iii), represents family or system of lines where  \lambda \in R .


 (a_1x+b_1y+c_1)+\lambda (a_2x+b_2y+c_2) =0 , represents family of lines.

(i) through the point of intersections  L_1    and  L_2 , if they intersect.

(ii) parallel to  L_1    and  L_2 , if they are parallel.

Examples for Corresponding Conditions:

 (i) \quad L_1: 2x-5y+2 =0, \quad L_2: 5x-4y-5=0, \quad L_3: (2x-5y+2)+\lambda(5x-4y+5)=0

  (ii)\quad L_1: 2x-5y+3 =0, \quad L_2: 54x-10y+1=0, \quad L_3: (2x-5y+3)+\lambda(4x-10y+1)=0


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