Exercise 6.4-Additional Questions - Chapter 6 Two Dimensional Analytical Geometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions Solved
Question 1.
Find the angle between the pair of straight lines given by $\left(a^2-3 b^2\right) x^2+8 a b x y+\left(b^2-3 a^2\right) y^2=0$.
Solution:
Angle between the lines $a x^2+2 h x y+h y^2=0$ is given by $\tan \theta=\frac{ \pm 2 \sqrt{h^2-a b}}{a+b}$
In this problem, $\tan \theta=\frac{ \pm 2 \sqrt{16 a^2 b^2-\left(a^2-3 b^2\right)\left(b^2-3 a^2\right)}}{a^2-3 b^2+b^2-3 a^2}$
$=\frac{ \pm 2 \sqrt{16 a^2 b^2-a^2 b^2+3 b^4+3 a^4-9 a^2 b^2}}{-2 a^2-2 b^2}$
$=\frac{ \pm 2 \sqrt{3 a^4+3 b^4+6 a^2 b^2}}{-2\left(a^2+b^2\right)}= \pm \sqrt{3}$
$\tan \theta=60^{\circ}$ [If we take the acute angle]

Question 2.
Show that $9 x^2+24 x y+16 y^2+21 x+28 y+6=0$ represents a pair of parallel straight lines and find the distance between them.
Solution:
$
9 x^2+24 x y+16 y^2+21 x+28 y+6=0
$
Here $\mathrm{a}=9.6$
$
\begin{aligned}
& \mathrm{b}=16 \\
& \mathrm{~g}=\frac{21}{2} \\
& \mathrm{f}=14 \\
& \mathrm{c}=6 \\
& \mathrm{~h}=12 \\
& \mathrm{~h}^2-\mathrm{ab}=(12)^2-9(16)=144-144=0
\end{aligned}
$
$\therefore$ The lines are parallel.
$
9 \mathrm{x}^2+24 \mathrm{xy}+16 \mathrm{y}^2=(3 \mathrm{x}+4 \mathrm{y})(3 \mathrm{x}+4 \mathrm{y})
$
Let $9 x^2+24 x y+16 y^2+21 x+28 y+6=(3 x+4 y+1)(3 x+4 y+m)$
Equating the coefficients of $x$ and constant term
$
\begin{aligned}
& 31+3 \mathrm{~m}=21 \\
& 1 \mathrm{~m}=6
\end{aligned}
$
Solving we get, $1=1$ or 6
$
m=6 \text { or } 1
$
$\therefore$ The separate equations are $3 x+4 y+1=0$ and $3 x+4 y+6=0$
The distance between the parallel lines are $\left|\frac{6-1}{\sqrt{9+16}}\right|=\frac{5}{5}=1$ unit.

Question 3.
If the equation $12 x^2-10 x y+2 y^2+14 x-5 y+c=0$ represents a pair of straight lines, find the value of c. Find the separate equations of the straight lines and also the angle between them.
Solution:
$
\begin{aligned}
& 12 x^2-10 x y-2 y^2+14 x-5 y+c=0 \\
& a x^2+2 h x y+b y^2+2 g x+2 f y-c=0
\end{aligned}
$
Here $\mathrm{a}=12$,
$
\begin{aligned}
& \mathrm{b}=2 \\
& \mathrm{~g}=7 \\
& \mathrm{f}=5 / 2 \\
& \mathrm{c}=\mathrm{c} \\
& \mathrm{h}=-5 \\
& \mathrm{af}^2+\mathrm{bg}^2+\mathrm{ch}^2-2 \mathrm{fgh}-\mathrm{abc}=0 \text { is the condition } \\
& 12\left(\frac{25}{4}\right)+2(7)^2+c(-5)^2-2\left(\frac{-5}{2}\right)(7)(-5)-12(2)(c)=0 \\
& 75+98+25 c-175-24 c=0, c=2
\end{aligned}
$
The equation is $12 \mathrm{x}^2-10 \mathrm{y}+2 \mathrm{y}^2+14 \mathrm{x}-5 \mathrm{y}+2=0$
$
12 x^2-10 x y+2 y=(3 x-y)(4 x-2 y)
$
Let $12 x^2-10 y+2 y^2+14 x-5 y+2(3 x-y+1)(4 x-2 y+m)$
So that $41+3 \mathrm{~m}=14,-21-\mathrm{m}=-5$
On solving we get $l=\frac{1}{2}, m=4$
$\therefore$ The separate equations are $3 x-y+\frac{1}{2}=0 \Rightarrow 6 x-2 y+1=0$ and
$
\begin{aligned}
& 4 x-2 y+4=0 \Rightarrow 2 x-y+2=0 \\
& m_1=\frac{-6}{-2}=3 ; m_2=\frac{-2}{-1}=2 \\
& \therefore \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{3-2}{1+3.2}\right|=\frac{1}{7} \\
& \theta=\tan ^{-1}\left(\frac{1}{7}\right)
\end{aligned}
$

Question 4.
For what value of $\mathrm{k}$ does $12 \mathrm{x}^2+7 \mathrm{xy}+\mathrm{ky}^2+13 \mathrm{x}-\mathrm{y}+3=0$ represents a pair of straight lines? Also write the separate equations.
Solution:
$
\begin{aligned}
& 12 x^2+7 x y+k y^2+13 x-y+3=0 \\
& \mathrm{a}=12 \text {, } \\
& \mathrm{h}=\frac{7}{2} \\
&
\end{aligned}
$
$
\begin{aligned}
& \mathrm{f}=-\frac{1}{2} \\
& \mathrm{c}=3 \\
& \mathrm{af}^2+\mathrm{bg}^2+\mathrm{ch}^2-\mathrm{abc}-2 \mathrm{fgh}=0 \\
& 12\left(-\frac{1}{2}\right)^2+k\left(\frac{13}{2}\right)^2+3\left(\frac{7}{2}\right)^2-12(k)(3)-2\left(-\frac{1}{2}\right)\left(\frac{13}{2}\right)\left(\frac{7}{2}\right)=0 \\
& \frac{12}{4}+\frac{169 k}{4}+\frac{147}{4}-36 k+\frac{91}{4}=0 \\
& \Rightarrow 12+169 \mathrm{k}+147-144 \mathrm{k}+91=0 \\
& 25 \mathrm{k}=-250 \Rightarrow \mathrm{k}=-10
\end{aligned}
$
The equation is $12 x^2+7 x y-10 y^2+13 x-y+3=0$
To find separate equations: $12 \mathrm{x}^2+7 \mathrm{xy}-10 \mathrm{y}^2=(3 \mathrm{x}-2 \mathrm{y})(4 \mathrm{x}+5 \mathrm{y})$
Let $12 x^2+7 x y-10 y^2+13 x-y+3=0(3 x-2 y+1)(4 x+5 y+m)$
Equating the coefficient of $\mathrm{x} \Rightarrow 41+3 \mathrm{~m}=13$
Equating the coefficient of $\mathrm{y} \Rightarrow 51-2 \mathrm{~m}=-1$
(1) $\times 2 \Rightarrow 81+6 \mathrm{~m}=26$
(2) $\times 3 \Rightarrow 151-6 \mathrm{~m}=-3$
$231=23 \Rightarrow 1=1$
$4+3 \mathrm{~m}=13$
$
3 \mathrm{~m}=9 \Rightarrow \mathrm{m}=3
$
The separate equations are $3 x-2 y+1=0$ and $4 x+5 y+3=0$

Question 5.
Show that $3 x^2+10 x y+8 y^2+14 x+22 y+15=0$ represents a pair of straight lines and the angle between them is $\tan ^{-1}\left(\frac{2}{11}\right)$
Solution:
$
\begin{aligned}
& 3 x^2+10 x y+8 y^2+14 x+22 y+15=0 \\
& a=3 \\
& A=5 \\
& b=8 \\
& g=7 \\
& f=11 \\
& c=15
\end{aligned}
$
The condition is $a f^2+\mathrm{bg}^2+\mathrm{ch}^2-\mathrm{abc}-2 \mathrm{fgh}=0$
$
3(11)^2+8(7)^2+15(5)^2-(3)(8)(15)-2(11)(7)(5)=363+392+375-360-770=0
$
The angle between the pair of straight line is given by
$
\begin{aligned}
& \tan \theta=\frac{ \pm 2 \sqrt{h^2-a b}}{a+b}=\frac{ \pm 2 \sqrt{25-3(8)}}{3+8}= \pm \frac{2}{11} \\
& \tan \theta=\frac{2}{11} \Rightarrow \theta=\tan ^{-1}\left(\frac{2}{11}\right)
\end{aligned}
$