⇒Pair of Straight Line

Correct answer option is (C)

$\begin{array}{c}(x\cos \alpha +y\sin \alpha {)}^2=(x^2+y^2){\sin }^2\alpha \\ \therefore x^2{\cos }^2\alpha +y^2{\sin }^2\alpha +2xy\sin \alpha \cos \alpha \\ =x^2{\sin }^2\alpha +y^2{\sin }^2\alpha \\ \therefore x^2({\cos }^2\alpha -{\sin }^2\alpha )+2xy\sin \alpha \cos \alpha =0\end{array}$

This represents a pair of straight lines where $\mathrm{a}={\cos }^2\alpha -{\sin }^2\alpha ,\mathrm{h}=\sin \alpha \cos \alpha$ and $\mathrm{b}=0$

As lines are perpendicular, we get $a+b=0$

$\therefore {\cos }^2\alpha ={\sin }^2\alpha \Rightarrow \alpha =\frac{\pi }{4}$

Correct answer option is (A)

Given equation of pair of lines is

$\begin{array}{ll}& x^2+\lambda xy-y^2{\tan }^2\theta =0\\ \therefore & a=1,h=\frac{\lambda }{2},\mathrm{b}=-{\tan }^2\theta \\ \therefore & \tan 2\theta =\left|\frac{2\sqrt{{\mathrm{h}}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right|\\ & \Rightarrow \frac{2\tan \theta }{1-{\tan }^2\theta }=\left|\frac{2\sqrt{\frac{{\lambda }^2}{4}+{\tan }^2\theta }}{1-{\tan }^2\theta }\right|\\ & \Rightarrow \frac{{\lambda }^2}{4}+{\tan }^2\theta ={\tan }^2\theta \\ & \Rightarrow \lambda =0\end{array}$

Correct answer option is (A)

The slope of line $3x+2y-8=0$ is $m_1=\frac{-3}{2}$

Let $m$ be the slope of one of the lines making angle $\frac{\pi }{4}$ with $3x+2y-8=0$

$\begin{array}{rl}\therefore \tan \frac{\pi }{4}& =\left|\frac{m-m_1}{1+m_1}\right|\\ \Rightarrow 1& =\left|\frac{m-\left(\frac{-3}{2}\right)}{1+m\left(\frac{-3}{2}\right)}\right|\\ \Rightarrow 1& =\left|\frac{2m+3}{2-3m}\right|\end{array}$

Squaring on both sides, we get

$\begin{array}{rl}& (2-3m{)}^2=(2m+3{)}^2\\ & \Rightarrow 5m^2-24m-5=0\end{array}$

This is the auxiliary equation of two lines and their joint equation is obtained by putting

$\mathrm{m}=\frac{y}{x}$

The joint equation of the lines is

$\begin{array}{rl}& 5{\left(\frac{y}{x}\right)}^2-24\left(\frac{y}{x}\right)-5=0\\ & \text{ i.e., }5x^2+24xy-5y^2=0\end{array}$

Correct answer option is (A)

Given equation of pair of lines is

$\begin{array}{rl}& a{x}^2+(2\mathrm{a}+1)xy+2y^2=0\\ & \mathrm{A}=\mathrm{a},\mathrm{H}=\frac{2\mathrm{a}+1}{2},\mathrm{B}=2\end{array}$

Given condition,

$\begin{array}{ll}& {\mathrm{m}}_1=\frac{1}{{\mathrm{m}}_2}\\ \therefore & {\mathrm{m}}_1\cdot {\mathrm{m}}_2=1\\ & \text{ Product of slopes }=\frac{\mathrm{A}}{\mathrm{B}}=\frac{\mathrm{a}}{2}\\ \therefore & {\mathrm{m}}_1\cdot {\mathrm{m}}_2=1=\frac{\mathrm{a}}{2}\\ \therefore & \mathrm{a}=2\\ & \text{ Also, sum of slopes }=\frac{-2\mathrm{H}}{\mathrm{B}}=-\left(\frac{2\mathrm{a}+1}{2}\right)=\frac{-5}{2}\\ & \text{ Using }{({\mathrm{m}}_1+{\mathrm{m}}_2)}^2={\mathrm{m}}_1^2+{\mathrm{m}}_2^2+2{\mathrm{m}}_1{\mathrm{m}}_2\\ & {\left(\frac{-5}{2}\right)}^2={\mathrm{m}}_1^2+{\mathrm{m}}_2^2+2\times 1\\ \therefore & {\mathrm{m}}_1^2+{\mathrm{m}}_2^2=\frac{25}{4}-2\\ \therefore & {\mathrm{m}}_1^2+{\mathrm{m}}_2^2=\frac{17}{4}\end{array}$

Correct answer option is (A)

We have $3m^2+10mn+3n^2=4h^2$ ..... (1)

Given lines are $m{x}^2+2hx+n{y}^2=0$

Let $m_1$ and $m_2$ be the slopes of the lines

$\begin{array}{rl}& \therefore {\mathrm{m}}_1+{\mathrm{m}}_2=\frac{-2h}{\mathrm{n}}\text{ and }{\mathrm{m}}_1{\mathrm{m}}_2=\frac{\mathrm{m}}{\mathrm{n}}\\ & \begin{array}{rl}\text{ Now }{({\mathrm{m}}_1-{\mathrm{m}}_2)}^2& ={({\mathrm{m}}_1+{\mathrm{m}}_2)}^2-4{\mathrm{m}}_1{\mathrm{m}}_2\\ & ={\left(\frac{-2h}{\mathrm{n}}\right)}^2-\frac{4\mathrm{m}}{\mathrm{n}}=\frac{4{\mathrm{h}}^2-4\mathrm{mn}}{{\mathrm{n}}^2}=\frac{3{\mathrm{m}}^2+6\mathrm{mn}+3{\mathrm{n}}^2}{{\mathrm{n}}^2}\dots [\text{ From (1)] }\\ & =\frac{3(\mathrm{m}+\mathrm{n}{)}^2}{{\mathrm{n}}^2}\\ \therefore {\mathrm{m}}_1-{\mathrm{m}}_2=& \frac{\sqrt{3}(\mathrm{m}+\mathrm{n})}{\mathrm{n}}\end{array}\end{array}$

$\therefore$ Let $\theta$ be the required angle.

Now $\begin{array}{rl}\tan \theta & =\frac{|m_1-m_2|}{1+m_1{m}_2}\\ & =\frac{\left(\frac{\sqrt{3}(m+n)}{n}\right)}{(1+\frac{m}{n})}=\frac{\sqrt{3}(m+n)}{n}\times \frac{n}{m+n}=\sqrt{3}\Rightarrow \theta ={60}^{\circ }=\frac{{\pi }^c}{3}\end{array}$