Wehave, 
\(\underset { n\rightarrow \infty }{ lim } \cfrac { 1+2+3+..+n }{ { n }^{ 2 } } =\underset { n\rightarrow \infty }{ lim } \cfrac { 1 }{ { n }^{ 2 } } \times \left( \cfrac { n\left( n+1 \right) }{ 2 } \right) \)  \(\left[ \because 1+2+...+n=\cfrac { n(n+1) }{ 2 } \right] \)
= \(\underset { n\rightarrow \infty }{ lim } \cfrac { 1 }{ 2 } \left( 1+\cfrac { 1 }{ n } \right) =\cfrac { 1 }{ 2 } \) 

\({\tan}^{-1}\left( {2 \over 11} \right)\)

k = 2, 2x - y + 2 = 0, 6x - 2y + 1 = 0, \(\theta={\tan}^{-1}\left({1\over 7} \right)\)

90°

5x + 3y + 8 = 0

3x - 8y + 13= 0

x +y + 2 = 0

Let P (h, k) be a point on the locus, From the given information, we have h = ct and k = \(\frac{c}{t}\). To eliminate t, taking product of these two equations
(h)(k) = (ct)(\(\frac{c}{t}\)) ⇒ hk = c²
Therefore, the required locus is xy = c²

Given pair of lines is 2x²- 5xy - 7y² = 0
Here a = 2, 2h = -5 and b = -7
\(h=\frac{-5}{2}\)
If \(\theta\) is the acute angle between the lines, then 
\(\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}\)
\(=\pm\frac{2\sqrt{\frac{25}{4}-2(-7)}}{2-7}\)
\(=\pm\frac{2\sqrt{\frac{25}{4}+14)}}{-5}=\pm\frac{2\sqrt{81}}{2(5)}=\pm\frac{\sqrt{81}}{5}=\frac{9}{5}\)
\(\Rightarrow\tan\theta=\frac{9}{5}\)

12x + 5y = 7 and 12x + 5y + 7 = 0
Given parallel lines are 12x + 5y = 7 and 12x + 5y + 7 = 0
Here a = 12, b = 5, c₁ = -7  and c₂ = 7
Distance between parallel lines = \(\left| \frac { { c }_{ 1 }-{ c }_{ 2 } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \right| \)
= \(\left| \frac { -7-7 }{ \sqrt { { 12 }^{ 2 }+{ 5 }^{ 2 } } } \right| =\left| \frac { -14 }{ \sqrt { 169 } } \right| \)
Distance between parallel lines = \(\frac { 14 }{ 13 } \)