Exercise 6.5 - Chapter 6 Two Dimensional Analytical Geometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

E.X:6.5

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Question 1.
The equation of the locus of the point whose distance from y-axis is half the distance from origin is
(a) $x^2+3 y^2=0$
(b) $x^2-3 y^2=0$
(c) $3 x^2+y^2=0$
(d) $3 x^2-y^2=0$
Solution:
(c) $3 x^2+y^2=0$
Hint:
Given that $\mathrm{PA}=\backslash([\backslash \operatorname{frac}\{1\}\{2\} /$ latex $] \mathrm{OP}$

$2 \mathrm{PA}=\mathrm{OP}$
$4 \mathrm{PA}^2=\mathrm{OP}^2$
$4(x)^2=x^2+y^2 \Rightarrow 3 x^2-y^2=0$
Question 2.
Which of the following equation is the locus of (at ${ }^2, 2$ at) .....
(a) $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
(b) $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
(c) $x^2+y^2=a^2$
(d) $y^2=4 a x$
Solution:
(d) $y^2=4 \mathrm{ax}$
Hint:
$y^2=4 \mathrm{ax} \Rightarrow$ Equation that satisfies the given point (at $\left.{ }^2, 2 \mathrm{at}\right)$

Question 3.
Which of the following point lie on the locus of $3 x^2+3 y^2-8 x-12 y+17=0$ ?
(a) $(0,0)$
(b) $(-2,3)$
(c) $(1,2)$
(d) $(0,-1)$
Solution:
(c) $(1,2)$
Hint:
The point that satisfies the given equations $(0,0) \Rightarrow 17 \neq 0$
$
\begin{aligned}
& (-2,3) \Rightarrow 3(4)+3(9)+16-36+17 \neq 0 \\
& (1,2) \Rightarrow 3+3(4)-8(1)-12(2)+17 \\
& 32-32=0,0=0
\end{aligned}
$
Question 4.
If the point $(8,-5)$ lies on the locus $\frac{x^2}{16}-\frac{y^2}{25}=k$ then the value of $k$ is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(d) 3
$\operatorname{Sub}(x, y)=(8,-5) \Rightarrow \frac{64^4}{16}-\frac{25}{25}=k$ $4-1=k \Rightarrow k=3$
Question 5 .
Straight line joining the points $(2,3)$ and $(-1,4)$ passes through the point $(\alpha, \beta)$ if
(a) $\alpha+2 \beta=7$
(b) $3 \alpha+\beta=9$
(c) $\alpha+3 \beta=11$
(d) $3 \alpha+3 \beta=11$
Solution:
(c) $\alpha+3 \beta=11$
Hint:
Equation joining $(2,3),(-1,4)$
$
\begin{aligned}
& \frac{y-4}{3-4}=\frac{x+1}{2+1} \Rightarrow \frac{y-4}{-1}=\frac{x+1}{3} \\
& 3 \mathrm{y}-12=-\mathrm{x}-1 \Rightarrow \mathrm{x}+3 \mathrm{y}-11=0,(\alpha, \beta) \text { lies on it } \Rightarrow \alpha+3 \beta-11=0 .
\end{aligned}
$

Question 6.
The slope of the line which makes an angle $45^{\circ}$ with the line $3 x-y=-5$ are
(a) $1,-1$
(b) $[$ latex $]$ frac $\{1\}\{2\},-2 \backslash)$
(c) $1, \frac{1}{2}$
(d) $2,-\frac{1}{2}$
Solution:
(c) $1, \frac{1}{2}$
Hint:
Equation of line $3 x-y=-5, y=3 x+5, m_1=3$
Angle between two lines $\theta=45^{\circ}$
$
\begin{array}{c|c}
\tan ^{-1}\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=45^{\circ} \Rightarrow \frac{m_1-m_2}{1+m_1 m_2}= \pm \tan 45^{\circ} \\
\frac{3-m_2}{1+3 m_2}=1 & \frac{3-m_2}{1+3 m_2}=-1 \\
3-m_2=1+3 m_2 & 3-m_2=-1-3 m_2 \\
2=4 m_2 & 2 m_2=-4 \\
m_2=2 / 4 & m_2=-\frac{4}{2} \\
m_2=1 / 2 & m_2=-2
\end{array}
$
The slopes are $-2, \frac{1}{2}$

Question 6.
The slope of the line which makes an angle $45^{\circ}$ with the line $3 x-y=-5$ are
(a) $1,-1$
(b) $[$ latex $] \backslash$ frac $\{1\}\{2\},-2 \backslash$
(c) $1, \frac{1}{2}$
(d) $2,-\frac{1}{2}$
Solution:
(c) $1, \frac{1}{2}$
Hint:
Equation of line $3 x-y=-5, y=3 x+5, m_1=3$
Angle between two lines $\theta=45^{\circ}$
$
\tan ^{-1}\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=45^{\circ} \Rightarrow \frac{m_1-m_2}{1+m_1 m_2}= \pm \tan 45^{\circ}
$

Question 7.
Equation of the straight line that forms an isosceles triangle with coordinate axes in the I-quadrant with perimeter $4+2 \sqrt{2}$ is
(a) $x+y+2=0$
(b) $x+y-2=0$
(c) $x+y-\sqrt{2}=0$
(d) $x+y+\sqrt{2}=0$
Solution:
(b) $x+y-2=0$
Hint.
Let the sides be $\mathrm{x}, \mathrm{x}$

$
\begin{aligned}
& \therefore \mathrm{AB}=\sqrt{x^2+x^2}=x \sqrt{2}, \text { Perimeter }=4+2 \sqrt{2} \\
& 2 x+x \sqrt{2}=4+2 \sqrt{2} \\
& x(2+\sqrt{2})=2(2+\sqrt{2}) \\
& x, y \text { intercept }=2, \text { Equation of line is } \frac{x}{2}+\frac{y}{2}=1 \\
& x+y=2 \Rightarrow x+y-2=0
\end{aligned}
$
Question 8 .
The coordinates of the four vertices of a quadrilateral are $(-2,4),(-1,2),(1,2)$ and $(2,4)$ taken in order. The equation of the line passing through the vertex $(-1,2)$ and dividing the quadrilateral in the equal areas is
(a) $x+1=0$
(b) $x+y=1$
(c) $x+y+3=0$
(d) $x-y+3=0$
Solution:
(b) $x+y=1$
Hint:
This equation passes through $(-1,2)$
$
-1+2=1 \Rightarrow 1=1
$

Question 9.
The intercepts of the perpendicular bisector of the line segment joining $(1,2)$ and $(3,4)$ with coordinate axes are
(a) $5,-5$
(b) 5,5
(c) 5,3
(d) $5,-4$
Solution:
(b) 5,5
$
\begin{aligned}
& (1,2),(3,4) \Rightarrow \text { Mid Point }=\left(\frac{1+3}{2}, \frac{2+4}{2}\right)=(2,3) \\
& \text { Slope }=\frac{4-2}{3-1}=\frac{2}{2}=1 ; \text { Perpendicular Slope }=\frac{-1}{m}=-1
\end{aligned}
$
Equation of the perpendicular bisector is $y-y_1=m\left(x-x_1\right)$
$
\begin{aligned}
& y-3=-1(x-2) \Rightarrow y-3=-x+2 \Rightarrow x+y=5 \Rightarrow \frac{x}{5}+\frac{y}{5}=1 \\
& x, y \text { intercepts are }(5,5)
\end{aligned}
$
Question 10.
The equation of the line with slope 2 and the length of the perpendicular from the origin equal to $\sqrt{5}$ is
(a) $x+2 y=\sqrt{5}$
(b) $2 x+y=\sqrt{5}$
(c) $2 x+y=5$
(d) $x+2 y-5=0$
Solution:
(c) $2 \mathrm{x}+\mathrm{y}=5$
Hint: $y=2 x+c$
Perpendicular distance from origin $=\frac{c}{\sqrt{a^2+b^2}}=\sqrt{5}$

$
\frac{c}{\sqrt{1+4}}=\sqrt{5} \Rightarrow c=\sqrt{5} \cdot \sqrt{5}=5
$
The required line is $y=2 x+5 \Rightarrow 2 x-y+5=0$
Question 11.
A line perpendicular to the line $5 x-y=0$ forms a triangle with the coordinate axes. If the area of the triangle is 5 sq. units, then its equation is
(a) $x+5 y \pm 5 \sqrt{2}=0$
(b) $x-5 y \pm 5 \sqrt{2}=0$
(c) $5 x+y \pm 5 \sqrt{2}=0$
(d) $5 x-y \pm 5 \sqrt{2}=0$
Solution:
(a) $x+5 y \pm 5 \sqrt{2}=0$
Hint:

Equation of a line perpendicular to $5 x-y=0$ is
$
\begin{aligned}
& x+5 y=k, \frac{x}{k}+\frac{y}{\frac{k}{5}}=1 \\
& x, y \text { intercepts are } k, \frac{k}{5} \\
& \text { Area }=5 \\
& \frac{1}{2} k \times \frac{k}{5}=5, k^2=50, k^2= \pm 5 \sqrt{2}
\end{aligned}
$
Equation of line is $x+5 y \pm 5 \sqrt{2}=0$
Question 12 .
Equation of the straight line perpendicular to the line $x-y+5=0$, through the point of intersection the $y$-axis and the given line
(a) $x-y-5=0$
(b) $x+y-5=0$
(c) $x+y+5=0$
(d) $x+y+10=0$
Solution:
(b) $x+y-5=0$
Hint:
$
\mathrm{x}-\mathrm{y}+5=0 \Rightarrow \text { put } \mathrm{x}=0, \mathrm{y}=5
$
The point is $(0,5)$
Equation of a line perpendicular to $x-y+5=0$ is $x+y+k=0$
This passes through $(0,5)$
$
\begin{aligned}
& \mathrm{k}=-5 \\
& \mathrm{x}+7-5=0
\end{aligned}
$
Question 13.
If the equation of the base opposite to the vertex $(2,3)$ of an equilateral triangle is $x+y=2$, then the length of a side is .......
(a) $\sqrt{\frac{3}{2}}$
(b) 6
(c) $\sqrt{6}$
(d) $3 \sqrt{2}$
Solution:
$\sqrt{6}$
Hint:
In an equilateral $\Delta$ the perpendicular will bisects the base in to two equal parts. Length of the perpendicular drawn from $(2,3)$ to the line $\mathrm{x}+7-2=0$
$
\begin{aligned}
& =\left|\frac{2+3-2}{\sqrt{1+1}}\right|=|3 / \sqrt{2}| s \\
& \text { This is equal to } \frac{3}{\sqrt{2}}=\frac{\sqrt{3} a}{2} \\
& a=6 / \sqrt{6}=\sqrt{6}
\end{aligned}
$
Question 14.
The line $(p+2 q) x+(p-3 q) y=p-q$ for different values of $\mathrm{p}$ and $q$ passes through the point .....
(a) $\left(\frac{3}{2}, \frac{5}{2}\right)$
(b) $\left(\frac{2}{5}, \frac{2}{5}\right)$
(c) $\left(\frac{3}{5}, \frac{3}{5}\right)$
(d) $\left(\frac{2}{5}, \frac{3}{5}\right)$
Solution:
(d) $\left(\frac{2}{5}, \frac{3}{5}\right)$
Hint:
$
\begin{aligned}
& (\mathrm{p}+2 \mathrm{q}) \mathrm{x}+(\mathrm{p}-3 \mathrm{q}) \mathrm{y}=\mathrm{p}-\mathrm{q} \\
& \mathrm{px}+2 \mathrm{qx}+\mathrm{py}-3 \mathrm{qy}=\mathrm{p}-\mathrm{q} \\
& \mathrm{P}(\mathrm{x}+\mathrm{y})+\mathrm{q}(2 \mathrm{x}-3 \mathrm{y})=\mathrm{p}-\mathrm{q} \\
& \text { The fourth option } \mathrm{x}=2 / 5, \mathrm{y}=3 / 5 \\
& p\left(\frac{2}{5}+3 / 5\right)+q\left(\frac{4}{5}-\frac{9}{5}\right)=p(5 / 5)+q\left(\frac{-5}{5}\right) \\
& =\mathrm{p}-\mathrm{q}=\text { RHS }
\end{aligned}
$
Question 15.
The point on the line $2 x-3 y=5$ is equidistance from $(1,2)$ and $(3,4)$ is ...
(a) $(7,3)$
(b) $(4,1)$
(c) $(1,-1)$
(d) $(-2,3)$
Solution:
(b) $(4,1)$
Hint:
Let $(a, b)$ be on $2 \mathrm{x}-3 \mathrm{y}=5 \Rightarrow 2 \mathrm{a}-3 \mathrm{~b}=5$
It is equidistance from $(1,2)$ and $(3,4)$
$
\begin{aligned}
& \sqrt{(a-1)^2+(b-2)^2}=\sqrt{(a-3)^2+(b-4)^2} \\
& (a-1)^2+(b-2)^2=(a-3)^2+(6-4)^2 \\
& a^2-2 a+1+b^2-4 b+4=a^2-6 a+9+b^2-8 b+16 \\
& 4 a+4 b=20 \\
& 2 a+2 b=10 \\
& 2 a-3 b=5
\end{aligned}
$

$
\begin{aligned}
& 5 \mathrm{~b}=5 \\
& \mathrm{~b}=1 \therefore \mathrm{a}=4
\end{aligned}
$
$\therefore$ The point is $(4,1)$
Question 16.
The image of the point $(2,3)$ in the line $y=-x$ is
(a) $(-3,-2)$
(b) $(-3,2)$
(c) $(-2,-3)$
(d) $(3,2)$
Solution:
(a) $(-3,-2)$
$
\begin{aligned}
& \text { Hint: } \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-2\left(a x_1+b y_1+c\right)}{\sqrt{a^2+b^2}} \\
& \frac{x-2}{1}=\frac{y-3}{1}=\frac{-2(2+3)}{1+1} \\
& \frac{x-2}{1}=\frac{y-3}{1}=\frac{-105}{2} \\
& \mathrm{x}-2=-5, \mathrm{y}-3=-5 \\
& \mathrm{x}=-3, \mathrm{y}=-2 \\
& (-3,-2)
\end{aligned}
$

Question 17.
The length of $\perp$ from the origin to the line $\frac{x}{3}-\frac{y}{4}=1$ is
(a) $\frac{11}{5}$
(b) $\frac{5}{12}$
(c) $\frac{12}{5}$
(d) $-\frac{5}{12}$
Solution:
(c) $\frac{12}{5}$
Hint:
$
\begin{aligned}
& 4 \mathrm{x}-3 \mathrm{y}=12 \Rightarrow 4 \mathrm{x}-3 \mathrm{y}-12=0 \\
& d=\left|\frac{-12}{\sqrt{16+9}}\right|=\frac{12}{5}
\end{aligned}
$
Question 18.
The $y$-intercept of the straight line passing through $(1,3)$ and perpendicular to $2 x-3 y+1=0$ is
(a) $\frac{3}{2}$
(b) $\frac{9}{2}$
(c) $\frac{2}{3}$
(d) $\frac{2}{9}$
Solution:
(b) $\frac{9}{2}$
Hint:
Equation of a line perpendicular to $2 \mathrm{x}-3 \mathrm{y}+1=0$ is $3 \mathrm{x}+2 \mathrm{y}=\mathrm{k}$. It passes through $(1,3)$. $3+6=\mathrm{k} \Rightarrow \mathrm{k}=9,3 \mathrm{x}+2 \mathrm{y}=9$
To find $y$-intercept $x=0,2 y=9, y=9 / 2$

Question 19.
If the two straight lines $x+(2 k-7) y+3=0$ and $3 \mathrm{kx}+9 \mathrm{y}-5=0$ are perpendicular then the value of $\mathrm{k}$ is ......
$(a) k=3$
(b) $k=\frac{1}{3}$
(c) $k=\frac{2}{3}$
(d) $k=\frac{3}{2}$
Solution:
(a) $\mathrm{k}=3$
Hint.
$
\begin{aligned}
& x+(2 k-7) y+3=0 \Rightarrow m_1=\frac{-a}{b}=\frac{-1}{2 k-7} \\
& 3 k x+9 y-5=0, m_2=-\frac{3 k}{9}=-\frac{k}{3}
\end{aligned}
$
Since the lines are perpendicular $\mathrm{m}_1 \mathrm{~m}_2=-1$
$
\left(\frac{-1}{2 k-7}\right)\left(-\frac{k}{3}\right)=-1 \Rightarrow k=-3(2 k-7) \Rightarrow k=-6 k+21 \Rightarrow 7 k=21 \Rightarrow k=21 / 7=3
$
Question 20 .
If a vertex of a square is at the origin and its one side lies along the line $4 x+3 y-20=0$, then the area of the square is
(a) 20 sq. units
(b) 16 sq. units
(c) 25 sq. units
(d) 4 sq.units
Solution:
(b) 16 sq. units
Hint:
One side of a square $=$ Length of the perpendicular from $(0,0)$ to the line.
$=\left|\frac{20}{\sqrt{16+9}}\right|=\left|\frac{20}{5}\right|, a=4$. Area of square $=a^2=16$ sq. units
Question 21.
If the lines represented by the equation $6 x^2+41 x y-7 y^2=0$ make angles $\alpha$ and $\beta$ with $x$ - axis, then $\tan \alpha \tan \beta=$
(a) $-\frac{6}{7}$
(b) $-\frac{6}{7}$
(c) $-\frac{7}{6}$
(d) $\frac{7}{6}$
Solution:
(a) $-\frac{6}{7}$
Hint.
$
6 x^2+41 x y-7 y^2=0 \Rightarrow 6 x^2-x y+42 x y-7 y^2=0 \Rightarrow x(6 x-y)+7 y(6 x-y)=0
$

$
\begin{aligned}
& (\mathrm{x}+7 \mathrm{y})(6 \mathrm{x}-\mathrm{y})=0 \Rightarrow \mathrm{x}+7 \mathrm{y}=0,6 \mathrm{x}-\mathrm{y}=0 \\
& m_1=\frac{-1}{7}, m_2=\frac{-6}{-1}=6 \\
& \tan \alpha=-\frac{1}{7}, \tan \beta=6 \\
& \tan \alpha \tan \beta=(-1 / 7)(6) \\
& \tan \alpha \tan \beta=-6 / \mathrm{s}
\end{aligned}
$
Question 22 .
The area of the triangle formed by the lines $x^2-4 y^2=0$ and $x=a$ is
(a) $2 a^2$
(b) $\frac{\sqrt{3}}{2} a^2$
(c) $\frac{1}{2} a^2$
(d) $\frac{2}{\sqrt{3}} a^2$
Solution:
(c) $\frac{1}{2} a^2$
Hint:
$
\begin{aligned}
& \mathrm{x}^2-4 \mathrm{y}^2=0,(\mathrm{x}-2 \mathrm{y})(\mathrm{x}+2 \mathrm{y})=0 \Rightarrow \mathrm{x}-2 \mathrm{y}=0, \mathrm{x}+2 \mathrm{y}=0 \\
& \text { Area }=\frac{1}{2} b h=\frac{1}{2} \times a \times a=a^2 / 2
\end{aligned}
$

Question 23.
If one of the lines given by $6 x^2-x+4 x^2=0$ is $3 x+4 y=0$, then c equals to .....
(a) -3
(b) -1
(c) 3
(d) 1
Solution:
(a) -3
Hint.
$
6 x^2-x y+4 c y^2=0,3 x+4 y=0
$
The other line may be $(2 x+b y)$
$
\begin{aligned}
& (3 x+4 y)(2 x+b y)=6 x^2-x y+4 c y^2 \\
& 6 x^2+3 x b y+8 x y+4 b y^2=6 x^2-x y+4 c y^2 \\
& 6 x^2+x y(3 b+8)+4 b y^2=6 x^2-x y+4 c y^2
\end{aligned}
$
$
\begin{aligned}
& \text { paring, } 3 b+8=-1 \\
& 3 b=-9 \Rightarrow b=-3 \\
& 4 b=4 c \Rightarrow 4(-3)=4 c \\
& -12=4 c \Rightarrow c=-3
\end{aligned}
$

Question 24.
$\theta$ is acute angle between the lines $x^2-x y-6 y^2=0$ then $\frac{2 \cos \theta+3 \sin \theta}{4 \sin \theta+5 \cos \theta}$ is
(a) 1
(b) $-\frac{1}{9}$
(c) $\frac{5}{9}$
(d) $\frac{1}{9}$
Solution:
(c) $\frac{5}{9}$
Hint:
Hint: $x^2-x y-6 y^2=0, a=1, b=-6, h=-1 / 2$
$
\mathrm{Q}=\tan ^{-1}\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|
$
$
Q=\tan ^{-1}\left|\frac{2 \sqrt{\frac{1}{4}+6}}{-5}\right| \Rightarrow Q=\tan ^{-1}\left|\frac{2 \times 5 / 2}{5}\right|
$
$
\begin{aligned}
& Q=\tan ^{-1}(1)\left(\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right) \\
& Q=45^{\circ}
\end{aligned}
$
$
\frac{2 \cos \theta+3 \sin \theta}{4 \sin \theta+5 \cos \theta}=\frac{2 / \sqrt{2}+3 / \sqrt{2}}{\frac{4}{\sqrt{2}}+\frac{5}{\sqrt{2}}}=\frac{5 / \sqrt{2}}{\frac{9}{\sqrt{2}}}=5 / 9
$
Question 25.
The equation of one the line represented by the equation $x^2+2 x y \cot \theta-y^2=0$ is
(a) $x-y \cot \theta=0$
(b) $x+y \tan \theta=0$
(e) $x \cos \theta+y(\sin \theta+1)=0$
(d) $x \sin \theta+y(\cos \theta+1)=0$
Solution:

(d) $x \sin \theta+y(\cos \theta+1)=0$
Hint: $x^2+2 x y \cot \theta-y^2=0 \Rightarrow x^2+2 x y \cot \theta=y^2 \Rightarrow \frac{x^2}{y^2}+\frac{2 x}{y} \cot \theta=1$ By completing the squares
$
\begin{aligned}
& \left(\frac{x}{y}+\cot \theta\right)^2=1+\cot ^2 \theta \\
& \left(\frac{x}{y}+\cot \theta\right)^2=\operatorname{cosec}^2 \theta \\
& \frac{x}{y}+\cot \theta= \pm \operatorname{cosec} \theta
\end{aligned}
$