19) The path of an aeroplane is given by the equation 3x – 4y = 12. Represent the graph graphicall-
Also, show that the point (- ,-) lies on the graph.
Answers
Answer:
GiveN :
\begin{gathered} \rightsquigarrow \sf \: path \: \: of \: \: the \: \: aeroplane \: ;\: 3x - 4y = 12 \\ \end{gathered}
⇝pathoftheaeroplane;3x−4y=12
To FinD :
\begin{gathered} \rightsquigarrow \sf \: Check \: \: whether \: \: the \: \: aeroplane \: \: passes \: \: through \: \: the \: \: pt \: \: ( -4 \: , -6) \\ \end{gathered}
⇝Checkwhethertheaeroplanepassesthroughthept(−4,−6)
SolutioN :
\: \: \: \: \: \sf \: putting \: \: \: \: x= -4 \: \: \: \: \& \: \: \: \: y= -6puttingx=−4&y=−6
\mapsto \sf 3 \times ( - 4) - 4 \times ( - 6)↦3×(−4)−4×(−6)
\: = \sf \: 12=12
\begin{gathered} \therefore \sf \: As \: \: LHS \: \: is \: \: equal \: \: RHS \: \: we \: \: can \: \: say \: \: pt \: (-4,-6) \: exist \: \: in \: \: the \: \: line \\ \\ \\ \large{ : { \underline{ \boxed{ \mathfrak{Proved }}}} : }\end{gathered}
∴AsLHSisequalRHSwecansaypt(−4,−6)existintheline
:
Proved
:
ConcepT BoosteR :
\begin{gathered} \rightsquigarrow \sf For \: \: better \: \: understanding \: \: let's \: \: have \: \: a \: \: look \: \: on \: \: the \: \: graph \\ \end{gathered}
⇝Forbetterunderstandinglet
′
shavealookonthegraph
\underline { \underline{ \mathbb{ GRAPH \: \: IS \: \: ATTACHED \: \: IN \: \: PHOTO }}}
GRAPHISATTACHEDINPHOTO
