लेखक नवाब शाह की खीरे पर नमक मिर्च लगाता देखकर क्या सोच रहे थे
Answers
Answer:
We know that :-
\begin{gathered} \tt If \: slope \: of \:two \: lines \: are \: m_1 \: and \: m_2 then \: acute \: \: angle \: \theta \: between \: two \: lines \: is \: given \: as : \\ \\ \sf tan\theta = \bigg | \frac{m_1 -m_2 }{1 +m_1 m_2 } \bigg |\end{gathered}
Ifslopeoftwolinesarem
1
andm
2
thenacuteangleθbetweentwolinesisgivenas:
tanθ=
∣
∣
∣
∣
∣
1+m
1
m
2
m
1
−m
2
∣
∣
∣
∣
∣
We also know that :-
\begin{gathered} \tt slope (m) \: \: of \: line \: \: ax + by + c = 0 \: is : \\ \\ \boxed{ \sf m = \frac{ - a}{b} }\end{gathered}
slope(m)oflineax+by+c=0is:
m=
b
−a
\begin{gathered}\tt slope (m _1) \: \: of \: line \: \: y - \sqrt{3} x - 5 = 0 : \\ \\ \rightarrow \sf m _1 = - \: \frac{ - \sqrt{3} }{1} = \sqrt{3} \\ \\ \\ \tt slope (m _2) \: \: of \: line \: \: \sqrt{3} y - x + 6 = 0 : \\ \\ \sf \rightarrow m _2 = - \frac{ - 1}{ \sqrt{3} } = \frac{ 1}{ \sqrt{3}}\end{gathered}
slope(m
1
)ofliney−
3
x−5=0:
→m
1
=−
1
−
3
=
3
slope(m
2
)ofline
3
y−x+6=0:
→m
2
=−
3
−1
=
3
1
\begin{gathered}\sf \longrightarrow tan\theta = \bigg | \dfrac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 +(\sqrt{3} \times \frac{1}{ \sqrt{3}}) } \bigg | \\ \\ \sf \longrightarrow tan\theta = \bigg | \dfrac{ \frac{3 - 1}{ \sqrt{3} } }{1 +1} \bigg |\\ \\ \sf \longrightarrow tan\theta = \bigg | \dfrac{ \frac{2}{ \sqrt{3} } }{2} \bigg |\\ \\ \sf \longrightarrow tan\theta = \bigg | \frac{1}{ \sqrt{3} } \bigg |\\ \\ \sf \longrightarrow tan\theta = \frac{1}{ \sqrt{3} } \\ \\ \therefore \theta \: = 30 \degree\end{gathered}
⟶tanθ=
∣
∣
∣
∣
∣
1+(
3
×
3
1
)
3
−
3
1
∣
∣
∣
∣
∣
⟶tanθ=
∣
∣
∣
∣
∣
1+1
3
3−1
∣
∣
∣
∣
∣
⟶tanθ=
∣
∣
∣
∣
∣
2
3
2
∣
∣
∣
∣
∣
⟶tanθ=
∣
∣
∣
∣
∣
3
1
∣
∣
∣
∣
∣
⟶tanθ=
3
1
∴θ=30°
Angle between the lines y - √3x - 5 = 0 and √3y - x + 6 = 0 = 30°