$(A);$ If $(a_{1} x + b_{1} y + c_{1}) + (a_{2} x + b_{2} y + c_{2}) + a_{3} x + b_{3} y + c_{3}) = 0,$ then the lines $a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0, a_{3} x + b_{3} y + c_{3} = 0$ can not be parallel.
(R): If sum of three straight lines is identically $0,$ then they are either concurrent or parallel.

A and R are true and R is the correct explaination of A

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A and R are true and R not is the correct explaination of A

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A is true R is False

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A is False R is True

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Solution

The correct option is B A is False R is True
$(A) :$ If $(a_{1} x + b_{1} y + c_{1}) + (a_{2} x + b_{2} y + c_{2}) + (a_{3} x + b_{3} y + c_{3}) = 0$
Then $a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0$ and $a_{3} x + b_{3} y + c_{3} = 0$ $⟹$ they are either concurrent or parallel lines.
So, A is false.
$(R)$ is true (concept)
Hence, A is false and R is true.

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Similar questions

The condition for three lines $a_{1}$ x+ $b_{1}$ y+ $c_{1}$ = 0, $a_{2}$ x+ $b_{2}$ y+ $c_{2}$ = 0 and $a_{3}$ x+ $b_{3}$ y+ $c_{3}$ = 0 , to be concurrent is

$∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0$

The condition for three lines $a_{1} x + b_{1} y + c_{1} = 0$ , $a_{2} x + b_{2} y + c_{2} = 0$ and $a_{3} x + b_{3} y + c_{3} = 0$ , to be concurrent is

$∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0$

The condition for three lines $a_{1}$ x+ $b_{1}$ y+ $c_{1}$ = 0, $a_{2}$ x+ $b_{2}$ y+ $c_{2}$ = 0 and $a_{3}$ x+ $b_{3}$ y+ $c_{3}$ = 0 , to be concurrent is

$∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0$

Three lines given by $L_{1} : a_{1} x + b_{1} y + c_{1} = 0, L_{2} : a_{2} x + b_{2} y + c_{2} = 0 and L_{3} : a_{3} x + b_{3} y + c_{3} = 0$ are always concurrent given that

$∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0$

Three lines given by $4 : a_{1} x + b_{1} y + c = 0, L_{2} : a_{2} x + b_{2} y + c_{2} = 0 a n d L_{3} : a_{3} x + b_{3} y + c_{3} = 0$ are always concurrent given that

$∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$ =0

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(A); If (a1x+b1y+c1)+(a2x+b2y+c2)+a3x+b3y+c3)=0, then the lines a1x+b1y+c1=0,a2x+b2y+c2=0,a3x+b3y+c3=0 can not be parallel.(R): If sum of three straight lines is identically 0, then they are either concurrent or parallel.

The correct option is B A is False R is True(A): If (a1x+b1y+c1)+(a2x+b2y+c2)+(a3x+b3y+c3)=0Then a1x+b1y+c1=0,a2x+b2y+c2=0 and a3x+b3y+c3=0 ⟹ they are either concurrent or parallel lines.So, A is false.(R) is true (concept)Hence, A is false and R is true.