Question

The path of an aeroplane is given by the equation 3x – 4y = 12. Represent the graph graphically. Also, show that the point (- 4, – 6) lies on the graph. PLEASE TELL CORRECT ANSWER ​

Answer 1

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GiveN :

$\rightsquigarrow \sf \: path \: \: of \: \: the \: \: aeroplane \: ;\: 3x - 4y = 12$

To FinD :

$\rightsquigarrow \sf \: Check \: \: whether \: \: the \: \: aeroplane \: \: passes \: \: through \: \: the \: \: pt \: \: ( -4 \: , -6)$

SolutioN :

$\: \: \: \: \: \sf \: putting \: \: \: \: x= -4 \: \: \: \: \& \: \: \: \: y= -6$

$\mapsto \sf 3 \times ( - 4) - 4 \times ( - 6)$

$\: = \sf \: 12$

$\therefore \sf \: As \: \: LHS \: \: is \: \: equal \: \: RHS \: \: we \: \: can \: \: say \: \: pt \: (-4,-6) \: exist \: \: in \: \: the \: \: line \\ \\ \\ \large{ : { \underline{ \boxed{ \mathfrak{Proved }}}} : }$

ConcepT BoosteR :

$\rightsquigarrow \sf For \: \: better \: \: understanding \: \: let's \: \: have \: \: a \: \: look \: \: on \: \: the \: \: graph$

$\underline { \underline{ \mathbb{ GRAPH \: \: IS \: \: ATTACHED \: \: IN \: \: PHOTO }}}$

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HOPE THIS IS HELPFUL...

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