5. Given the linear equation 2x + 3y - 8 = 0, write another linear equation in two variables x and y such that the geometric representation of the pair so formed is
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines (NCERT)
Answers
Answer:
answer is 1
intersecting lines
Step-by-step explanation:
Take X,Y-axes as shown in attached figure.
Suppose that the particle strikes the plane at a point P with coordinate (x,y). Consider the motion between A and P.
\begin{gathered} \sf{ \red{Motion~ in~ x-direction:}} \\ \: \: \: \: \: \: \: \: \: \sf{ \blue{Initial~ velocity = u}} \\ \: \ \: \: \: \sf{ \blue{Acceleration= 0}} \\ \: \: \: \: \: \: \: \: \: \: \: \ \: \sf{ \blue{x = ut}} \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i)\end{gathered}
Motion in x−direction:
Initial velocity=u
Acceleration=0
x=ut...(i)
\begin{gathered} \sf{ \red{Motion~ in~ y-direction:}} \\ \: \: \: \: \: \: \: \: \: \sf{ \blue{Initial~ velocity = 0}} \\ \: \ \: \: \: \sf{ \blue{Acceleration= g}} \\ \: \: \: \: \: \: \: \: \: \: \: \ \: \sf{ \blue{y = \frac{1}{2} {gt}^{2} }} \: \: \: \: \: ....(ii)\end{gathered}
Motion in y−direction:
Initial velocity=0
Acceleration=g
y=
2
1
gt
2
....(ii)
\maltese \: \sf{Eliminating \: \: t \: \: from \: (i) \: and \: (ii)}✠Eliminatingtfrom(i)and(ii)
\begin{gathered} \longrightarrow\: \: \: \: \: \: \: \: \: \sf {y = \frac{1}{2} g \frac{ {x}^{2} }{ {u}^{2} } } \\ \\ \: \: \: \: \: \: \: \: \: \: \longrightarrow\sf{ \pink{y = \tan\theta}}\end{gathered}
⟶y=
2
1
g
u
2
x
2
⟶y=tanθ
Thereafter,
\begin{gathered} \implies \sf { \frac{ {gx}^{2} }{ {2u}^{2} }} = x \tan\theta \: \\ \\ \implies\sf \bold{x = \frac{2 {u}^{2} \tan \theta }{g}}\end{gathered}
⟹
2u
2
gx
2
=xtanθ
⟹x=
g
2u
2
tanθ
\sf{Clearly \: the \: point \: P \: corresponds \: to,}ClearlythepointPcorrespondsto, \begin{gathered} \\ ~~~~~~~~~~ \sf{\: x = \: \frac{2 {u}^{2} \: \tan \theta }{g} }\end{gathered}
x=
g
2u
2
tanθ
then,
\sf{y = x \tan \theta = \frac{2 {u}^{2} { \tan}^{2} \theta }{g} }y=xtanθ=
g
2u
2
tan
2
θ
\begin{gathered} \sf{The \:distance \: \: AP = } \: l = \sqrt{ {x}^{2} + {y}^{2} } \\ \\ \: \: \: \: \: \: \: \implies \sf{\frac{2 {u}^{2} }{g} tan \theta \sqrt{1 + { \tan}^{2} \theta } } \\ \\ \: : \implies \sf{ \pink{ \frac{2 {u}^{2} }{g} \tan \theta \sec \theta }}\end{gathered}
ThedistanceAP=l=
x
2
+y
2
⟹
g
2u
2
tanθ
1+tan
2
θ
:⟹
g
2u
2
tanθsecθ
Take X,Y-axes as shown in attached figure.
Suppose that the particle strikes the plane at a point P with coordinate (x,y). Consider the motion between A and P.
\begin{gathered} \sf{ \red{Motion~ in~ x-direction:}} \\ \: \: \: \: \: \: \: \: \: \sf{ \blue{Initial~ velocity = u}} \\ \: \ \: \: \: \sf{ \blue{Acceleration= 0}} \\ \: \: \: \: \: \: \: \: \: \: \: \ \: \sf{ \blue{x = ut}} \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i)\end{gathered}
Motion in x−direction:
Initial velocity=u
Acceleration=0
x=ut...(i)
\begin{gathered} \sf{ \red{Motion~ in~ y-direction:}} \\ \: \: \: \: \: \: \: \: \: \sf{ \blue{Initial~ velocity = 0}} \\ \: \ \: \: \: \sf{ \blue{Acceleration= g}} \\ \: \: \: \: \: \: \: \: \: \: \: \ \: \sf{ \blue{y = \frac{1}{2} {gt}^{2} }} \: \: \: \: \: ....(ii)\end{gathered}
Motion in y−direction:
Initial velocity=0
Acceleration=g
y=
2
1
gt
2
....(ii)
\maltese \: \sf{Eliminating \: \: t \: \: from \: (i) \: and \: (ii)}✠Eliminatingtfrom(i)and(ii)
\begin{gathered} \longrightarrow\: \: \: \: \: \: \: \: \: \sf {y = \frac{1}{2} g \frac{ {x}^{2} }{ {u}^{2} } } \\ \\ \: \: \: \: \: \: \: \: \: \: \longrightarrow\sf{ \pink{y = \tan\theta}}\end{gathered}
⟶y=
2
1
g
u
2
x
2
⟶y=tanθ
Thereafter,
\begin{gathered} \implies \sf { \frac{ {gx}^{2} }{ {2u}^{2} }} = x \tan\theta \: \\ \\ \implies\sf \bold{x = \frac{2 {u}^{2} \tan \theta }{g}}\end{gathered}
⟹
2u
2
gx
2
=xtanθ
⟹x=
g
2u
2
tanθ
\sf{Clearly \: the \: point \: P \: corresponds \: to,}ClearlythepointPcorrespondsto, \begin{gathered} \\ ~~~~~~~~~~ \sf{\: x = \: \frac{2 {u}^{2} \: \tan \theta }{g} }\end{gathered}
x=
g
2u
2
tanθ
then,
\sf{y = x \tan \theta = \frac{2 {u}^{2} { \tan}^{2} \theta }{g} }y=xtanθ=
g
2u
2
tan
2
θ
\begin{gathered} \sf{The \:distance \: \: AP = } \: l = \sqrt{ {x}^{2} + {y}^{2} } \\ \\ \: \: \: \: \: \: \: \implies \sf{\frac{2 {u}^{2} }{g} tan \theta \sqrt{1 + { \tan}^{2} \theta } } \\ \\ \: : \implies \sf{ \pink{ \frac{2 {u}^{2} }{g} \tan \theta \sec \theta }}\end{gathered}
ThedistanceAP=l=
x
2
+y
2
⟹
g
2u
2
tanθ
1+tan
2
θ
:⟹
g
2u
2
tanθsecθ