Exercise 6.2 - Chapter 6 Two Dimensional Analytical Geometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

$\operatorname{Ex} 6.2$
Question 1.
Find the equation of the lines passing through the point $(1,1)$
(i) With y-intercept (- 4)
(ii) With slope 3
(iii) And (-2, 3)
(iv) And the perpendicular from the origin makes an angle $60^{\circ}$ with $x$-axis.
Solution:
(i) Given y intercept $=-4$,
Let $\mathrm{x}$ intercept be a
Now equation of the lines is $\frac{x}{a}+\frac{y}{-4}=1$.
It passes through $(1,1) \Rightarrow \frac{1}{a}-\frac{1}{4}=1, \frac{4-a}{4 a}=1$
$\Rightarrow 4-a=4 a \Rightarrow 4=4 a+a$
(i.e) $5 a=4 \Rightarrow a=4 / 5$
So the equation of the line is $\frac{x}{4 / 5}+\frac{y}{-4}=1$ (i.e.) $\frac{5 x}{4}-\frac{y}{4}=1$
(i.e) $\frac{5 x-y}{4}=1 \Rightarrow 5 x-y=4$ (or) $y=5 x-4$
(ii) Slope $\mathrm{m}=3$, passing through $\left(\mathrm{x}_1, \mathrm{y}_1\right)=(1,1)$
Equation of the line is $y-y_1=m\left(x-x_1\right)$
(i.e) $y-1=3(x-1) \Rightarrow y-1=3 x-3 \Rightarrow 3 x-y=2$

(iii) Passing through $(1,1)$ and $(-2,3)$
Equation of the line passing through 2 points is $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$
$
\begin{aligned}
& \text { (i.e) } \quad \frac{y-1}{3-1}=\frac{x-1}{-2-1} \text {, (i.e.) } \frac{y-1}{2}=\frac{x-1}{-3} \\
& -3(y-1)=2(x-1) \Rightarrow-3 y+3=2 x-2 \\
& 5=2 x+3 y \\
& 2 x+3 y=5
\end{aligned}
$
(iv) $\mathrm{P}=$ Distance between $(0,0)$ and $(1,1)=\sqrt{(0-1)^2+(0-1)^2}=\sqrt{1+1}=\sqrt{2}, \alpha=60^{\circ}$ Equation of the line is $x \cos \alpha+y \sin \alpha=p$
$
\begin{aligned}
& \text { (i:e) } x \cos 60^{\circ}+y \sin 60^{\circ}=\sqrt{2} \\
& x\left(\frac{1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=\sqrt{2} \Rightarrow \frac{x+\sqrt{3} y}{2}=\sqrt{2} \\
& \Rightarrow x+\sqrt{3} y=2 \sqrt{2}
\end{aligned}
$
Question 2.
If $\mathrm{P}(\mathrm{r}, \mathrm{c})$ is mid point of a line segment between the axes, then show that $\frac{x}{r}+\frac{y}{c}=2$.
Solution:
$\mathrm{P}(\mathrm{r}, \mathrm{c})$ is the mid point of $A B$.
$\Rightarrow \mathrm{A}=(2 \mathrm{r}, 0)$ and $\mathrm{B}=(0,2 \mathrm{c})$
(i.e) $\mathrm{x}$ intercept $=2 \mathrm{r}$ and
$\mathrm{y}$ intercept $=2 \mathrm{c}$.
Equation of the line is $\frac{x}{2 r}+\frac{y}{2 c}=1$
(i.e) $\frac{1}{2}\left(\frac{x}{r}+\frac{y}{c}\right)=1 \Rightarrow \frac{x}{r}+\frac{y}{c}=2$.

Question 3.
Find the equation of the line passing through the point $(1,5)$ and also divides the co-ordinate axes in the ratio $3: 10$.
Solution:
Let $\mathrm{x}$ intercept be $3 \mathrm{a}$ and $\mathrm{y}$ intercept be $10 \mathrm{a}$
Equation of the line is $\frac{x}{3 a}+\frac{y}{10 a}=1$
The line passes through $(1,5)$
$
\begin{gathered}
\Rightarrow \frac{1}{3 a}+\frac{5}{10 a}=1 \Rightarrow \frac{1}{3 a}+\frac{1}{2 a}=1 \\
\frac{2+3}{6 a}=1 \Rightarrow 6 a=5 \rightarrow \text { (1) }
\end{gathered}
$
So the equation of the line is $\frac{x}{3 a}+\frac{y}{10 a}=1 \Rightarrow \frac{10 x+3 y}{30 a}=1$
$
\Rightarrow 10 x+3 y=30 a
$
(i.e) $10 x+3 y=5(6 a)=5(5)=25$ [from (1)]
$
\therefore 10 x+3 y=25
$

Question 4.
If $p$ is length of perpendicular from origin to the line whose intercepts on the axes are $a$ and $b$, then show that $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$
Solution:
$
\begin{aligned}
& \text { (i.e) } \frac{b x+a y}{a b}=1 \Rightarrow b x+a y=a b \\
& \Rightarrow b x+a y-a b=0
\end{aligned}
$
$p=$ The length of the perpendicular from the origin to (1)
$
\begin{aligned}
& \Rightarrow p= \pm \frac{-a b}{\sqrt{a^2+b^2}} \\
& \text { (i.e) } p=\frac{a b}{\sqrt{a^2+b^2}}
\end{aligned}
$
Squaring on both sides $p^2=\frac{a^2 b^2}{a^2+b^2}$
$
\Rightarrow \frac{1}{p^2}=\frac{a^2+b^2}{a^2 b^2}=\frac{1}{b^2}+\frac{1}{a^2} \Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}
$
Question 5.
The normal boiling point of water is $100^{\circ} \mathrm{C}$ or $212^{\circ} \mathrm{F}$ and the freezing point of water is $0^{\circ} \mathrm{C}$ or $32^{\circ} \mathrm{F}$.
(i) Find the linear relationship between $\mathrm{C}$ and $\mathrm{F}$
(ii) Find the value of $\mathrm{C}$ for $98.6^{\circ} \mathrm{F}$ and
(iii) The value of $\mathrm{F}$ for $38^{\circ} \mathrm{C}$.
Solution:
Given when $\mathrm{C}=100, \mathrm{~F}=212$ and when $\mathrm{C}=0, \mathrm{~F}=32$

(i.e) $\left(\dot{x}_1, y_1\right)=(100,212)$ and $\left(x_2, y_2\right)=(0,32)$ and $(x, y)=(\mathrm{C}, \mathrm{F})$
(i) The equation of the line passing through $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$ (i.e) $\frac{\mathrm{F}-212}{32-212}=\frac{\mathrm{C}-100}{0-100} \Rightarrow \frac{\mathrm{F}-212}{-180}=\frac{\mathrm{C}-100}{-100}$ By dividing the denominators by $-20, \frac{\mathrm{F}-212}{9}=\frac{\mathrm{C}-100}{5}$
$
\Rightarrow \quad \begin{array}{r}
5 \mathrm{~F}-1060=9 \mathrm{C}-900 \\
5 \mathrm{~F}=9 \mathrm{C}-900+1060
\end{array}
$
(i.e) $5 \mathrm{~F}=9 \mathrm{C}+160$
$
\mathrm{F}=\frac{9 \mathrm{C}+160}{5}=\frac{9}{5} \mathrm{C}+32
$
Again $5 \mathrm{~F}-1060=9 \mathrm{C}-900$
(i.e) $9 \mathrm{C}-900=5 \mathrm{~F}-1060$
$9 \mathrm{C}=5 \mathrm{~F}-1060+900$
$9 \mathrm{C}=5 \mathrm{~F}-160$
$\mathrm{C}=\frac{5 \mathrm{~F}-160}{9}=\frac{5(\mathrm{~F}-32)}{9}$
$\mathrm{C}=\frac{5 \mathrm{~F}-160}{9}=\frac{5(\mathrm{~F}-32)}{9}$
$\mathrm{C}=\frac{5}{9}(\mathrm{~F}-32)$
(ii) $\mathrm{So} \mathrm{F}=\frac{9}{5} \mathrm{C}+32(\mathrm{OR}) \mathrm{C}=\frac{5}{9}(\mathrm{~F}-32)$
When $\mathrm{F}=98.6^{\circ}$ to find $\mathrm{C}$
$
\mathrm{C}=\frac{5}{9}(\mathrm{~F}-32)
$
When $\mathrm{F}^{\prime}=98.6, c=\frac{5}{9}(98.6-32)=\frac{5}{9}(66.6)=37^{\circ}$
(iii) When $\mathrm{C}=38^{\circ}$, To find $\mathrm{F}$
$
\begin{aligned}
& F=\frac{9}{5}(38)+32=9 \times 7.6+32=68.4+32 \\
& F=100.4^{\circ}
\end{aligned}
$

Question 6.
An object was launched from a place $P$ in constant speed to hit a target. At the 15 th second it was $1400 \mathrm{~m}$ away from the target and at the 18 th second $800 \mathrm{~m}$ away. Find
(i) The distance between the place and the target
(ii) The distance covered by it in 15 seconds,
(iii) Time taken to hit the target.
Solution:
Taking time $=\mathrm{x}$ and distance $=\mathrm{y}$
We are given at $\mathrm{x}=15, \mathrm{y}=1400$ and at $\mathrm{x}=18, \mathrm{y}=800$
The equation of the line passing through $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$
(i.e) $\frac{y-1400}{800-1400}=\frac{x-15}{18-15}$
(i.e) $\frac{y-1400}{-600}=\frac{x-15}{3}$
(Denominator divided by 3 ) we get $\frac{y-1400}{-200}=x-15$
(i.e) $y-1400=-200 x+3000$
$\Rightarrow 200 x=1400-y+3000 \Rightarrow x=\frac{1400-y}{200}+\frac{3000}{200}$
(i.e) $x=\frac{1400-y}{200}+15$
Taking $x=$ Time and $y=$ distance $(D)$ we get $T=\frac{1400-\mathrm{D}}{200}+15$
(i) Let $\mathrm{T}=0$ in (1)
$
0=\frac{1400-D}{200}+15 \Rightarrow 1400-\mathrm{D}+3000=0 \Rightarrow D=4400 \mathrm{~m}
$
(ii) Let $\mathrm{T}=15$ in $(1)$
$
15=\frac{1400-D}{200}+15 \Rightarrow 3000=1400-D+3000
$

(i.e) $1400-\mathrm{D}=0 \Rightarrow d=1400 \mathrm{~m}$
So the distance covered in 15 seconds $=4400-1400=3000 \mathrm{~m}$
(iii) To find $\mathrm{T}$ at $\mathrm{D}=0$
$
\mathrm{T}=\frac{1400-0}{200}+15 \Rightarrow \mathrm{T}=7+15=22 \text { seconds }
$
Question 7.
Population of a city in the years 2005 and 2010 are 1,35,000 and 1,45,000 respectively. Find the approximate population in the year 2015. (assuming that the growth of population is constant).
Solution:
Taking year as $\mathrm{x}$ and population as $\mathrm{y}$

We are given when $x=2005$, $y=1,35,000$ and
when $x=2010$,
$\mathrm{y}=1,45,000$
(i.e) $\left(x_1, y_1\right)=(2005,135000),\left(x_2, y_2\right)=(2010,145000)$
The equation of the line passing through $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$
(i.e) $\frac{y-135000}{145000-135000}=\frac{x-2005}{2010-2005}$
(i.e) $\quad \frac{y-135000}{10000}=\frac{x-2005}{5}$
$\frac{5}{10000}(y-135000)=x-2005$
(i.e) $\frac{1}{2000}(y-135000)=x-2005$
$\mathrm{y}-135000=2000(\mathrm{x}-2005)$
$\mathrm{y}=2000(\mathrm{x}-2005)+135000$
At $x=2015, y=2000(2015-2005)+135000$
(i.e) $y=2000(10)+135000=20000+135000=1,55,000$
The approximate population in the year 2015 is $1,55,000$

Question 8.
Find the equation of the line, if the perpendicular drawn from the origin makes an angle $30^{\circ}$ with $\mathrm{x}$ - axis and its length is 12 .
Solution:
The equation of the line is $\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{p}$
Here $\alpha=30^{\circ}, \cos \alpha=\cos 30^{\circ}=\frac{\sqrt{3}}{2} ; \sin \alpha=\sin 30^{\circ}=1 / 2 ; \mathrm{P}=12$.
So equation of the line is $x \frac{\sqrt{3}}{2}+y \frac{1}{2}=12$
(i.e) $\sqrt{3} x+y=12 \times 2=24 \Rightarrow \sqrt{3} x+y-24=0$
Question 9.
Find the equation of the straight lines passing through $(8,3)$ and having intercepts whose sum is 1 .
Solution:
Given sum of the intercepts $=1 \Rightarrow$ when $\mathrm{x}$ intercept $=\mathrm{a}$ then $\mathrm{y}$ intercept $=1-\mathrm{a}$
Equation of the line is $\frac{x}{a}+\frac{y}{1-a}=1$
The line passes through $(8,3) \Rightarrow \frac{8}{a}+\frac{3}{1-a}=1$
$
8(1-a)+3 a=a(1-a)
$

$
\begin{aligned}
& 8-8 a+3 a=a-a^2 \\
& \quad a^2-6 a+8=0 \\
& (a-2)(a-4)=0 \Rightarrow a=2 \text { or } 4
\end{aligned}
$
1. When $a=2$ equation of the line is $\frac{x}{2}+\frac{y}{1-2}=1 \Rightarrow \frac{x}{2}-y=1 \Rightarrow x-2 y=2$
2. When $a=4$ equation of the line is $\frac{x}{4}+\frac{y}{1-4}=1 \Rightarrow \frac{x}{4}-\frac{y}{3}=1 \Rightarrow 3 x-4 y=12$
Question 10.
Show that the points $(1,3),(2,1)$ and $\left(\frac{1}{2}, 4\right)$ are collinear, by using (i) Concept of slopi
(ii) Using a straight line and (iii) Any other method.
Solution:
Let the given points be $\mathrm{A}(1,3)$, $\mathrm{B}(2,1)$, and $\mathrm{C}\left(\frac{1}{2}, 4\right)$
(i) Slope of $\mathrm{AB}=\frac{1-3}{2-1}=\frac{-2}{1}=-2=m_1$
Slope of $\mathrm{BC}=\frac{4-1}{1 / 2-2}=\frac{3}{-3 / 2}=-2=m_2$
Slope of $\mathrm{AB}=$ Slope of $\mathrm{BC} \Rightarrow \mathrm{AB}$ parallel to $\mathrm{BC}$ but $\mathrm{B}$ is a common point $\Rightarrow$ The points $A, B, C$ are collinear.
(ii) Equation of the line passing through $\mathrm{A}$ and $\mathrm{B}$ is $\frac{y-1}{3-1}=\frac{x-2}{1-2} \Rightarrow \frac{y-1}{2}=\frac{x-2}{-1}$ $1-y=2 x-4$
$2 x+y=5$
Substituting C $\left(\frac{1}{2}, 4\right)$ in (1)
we get LHS $=2\left(\frac{1}{2}\right)+4=1+4=5=$ RHS
$C$ is a point on $A B$
$\Rightarrow$ The points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ lie on a line

$\Rightarrow$ The points A, B, C are collinear
(iii) Area of $\begin{aligned} \triangle \mathrm{ABC} & =\frac{1}{2}\left(x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right) \\ & =\frac{1}{2}\left\{1(1-4)+2(4-3)+\frac{1}{2}(3-1)\right\}=\frac{1}{2}(-3+2+1)=0\end{aligned}$
$\Rightarrow$ The points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are collinear.
Question 11.
A straight line is passing through the point $\mathrm{A}(1,2)$ with slope $\frac{5}{12}$. Find points on the line which are 13 units away from $A$.
Solution:

Equation of the line in parametric form is $\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=r$
Here $\left(x_1, y_1\right)=(1,2), r=13, m=\tan \theta=5 / 12$
$
\sin \theta=5 / 13, \cos \theta=12 / 13
$
$
\frac{x-1}{12 / 13}=\frac{y-2}{5 / 13}= \pm 13 \text {, (i.e) } \frac{13(x-1)}{12}=\frac{13(y-2)}{5}= \pm 13
$
$(\div$ by 13$) \frac{x-1}{12}=\frac{y-2}{5}= \pm 1$
$
\begin{array}{c|l}
\frac{x-1}{12}=1 \Rightarrow x=12+1=13 & \frac{x-1}{12}=-1 \Rightarrow x=-12+1=-11 \\
\frac{y-2}{5}=+1 \Rightarrow y=5+2=7 & \frac{y-2}{5}=-1 \Rightarrow y=-5+2=-3
\end{array}
$
So the points are $(13,7)$ or $(-11,-3)$
Question 12
A $150 \mathrm{~m}$ long train is moving with constant velocity of $12.5 \mathrm{~m} / \mathrm{s}$. Find
(i) The equation of the motion of the train,
(ii) Time taken to cross a pole,
(iii) The time taken to cross the bridge of length $850 \mathrm{~m}$ is?
Solution:
(i) Now $\mathrm{m}=\frac{y}{x}=12.5 \mathrm{~m} /$ second,
The equation of the line is $\mathrm{y}=\mathrm{mx}+\mathrm{c} \ldots$ (1)
Put $\mathrm{c}=-150, \mathrm{~m}=12.5 \mathrm{~m}$
The equation of motion of the train is $\mathrm{y}=12.5 \mathrm{x}-150$

(ii) To find the time taken to cross a pole we take $y=0$ in (1)
$
\begin{aligned}
& \Rightarrow 0=12.5 \mathrm{x}-150 \Rightarrow 12.5 \mathrm{x}=150 \\
& x=\frac{150}{12.5}=12 \text { seconds }
\end{aligned}
$
(iii) When $\mathrm{y}=850$ in (1)
$
\begin{aligned}
& 850=12.5 \mathrm{x}-150 \Rightarrow 12.5 \mathrm{x}=850+150=1000 \\
& \Rightarrow x=\frac{1000}{12.5}=80 \text { seconds }
\end{aligned}
$
Question 13.
A spring was hung from a hook in the ceiling. A number of different weights were attached to the spring to make it stretch, and the total length of the spring was measured each time shown in the following table.

(i) Draw a graph showing the results.
(ii) Find the equation relating the length of the spring to the weight on it.
(iii) What is the actual length of the spring.
(iv) If the spring has to stretch to $9 \mathrm{~cm}$ long, how much weight should be added?
(v) How long will the spring be when 6 kilograms of weight on it?
Solution:
Taking weight $(\mathrm{kg})$ as $\mathrm{x}$ values and length $(\mathrm{cm})$ as $\mathrm{y}$ values we get $\left(\mathrm{x}_1, \mathrm{y}_1\right)=(2,3),\left(\mathrm{x}_2, \mathrm{y}_2\right)=(4,4)$

(I)

The equation of the line passing through the above two points is
$
\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1} \text { (i.e) } \frac{y-3}{4-3}=\frac{x-2}{4-2}
$
(ii) $\frac{y-3}{1^*}=\frac{x-2}{2} \Rightarrow 2 y-6=x-2$
i.e. $x-2 y=-4 \Rightarrow x-2 y+4=0$
(iii) When $\mathrm{x}=0,2 \mathrm{y}=4 \Rightarrow \mathrm{y}=2 \mathrm{~cm}$
(iv) When $\mathrm{y}=9 \mathrm{~cm}, \mathrm{x}-18=-4$
$\mathrm{x}=-4+18=14 \mathrm{~kg}$
(v) When $x=6(\mathrm{~kg}), 6-2 \mathrm{y}=-4,-2 \mathrm{y}=-4-6=-10$
$\Rightarrow 2 \mathrm{y}=10 \Rightarrow \mathrm{y}=10 / 2=5 \mathrm{~cm}$.
Question 14.
A family is using Liquefied petroleum gas (LPG) of weight $14.2 \mathrm{~kg}$ for consumption. (Full weight 29.5 $\mathrm{kg}$ includes the empty cylinders tare weight of $15.3 \mathrm{~kg}$.). If it is use with constant rate then it lasts for 24 days. Then the new cylinder is replaced
(i) Find the equation relating the quantity of gas in the cylinder to the days.
(ii) Draw the graph for first 96 days.
Solution:

(i) Let $x$ represent the number of days.
$y$ represent the weight of gas

$\therefore$ Equation is $\frac{x}{24}+\frac{y}{14.2}=1 \Rightarrow \frac{y}{14.2}=1-\frac{x}{24}$
$
\begin{aligned}
& y=\frac{-14.2 x}{24}+14.2 \\
& y=-\frac{71 x}{120}+14.2,0 \leq x \leq 24
\end{aligned}
$

(ii)

Question 15 .
In a shopping mall there is a hall of cuboid shape with dimension $800 \times 800 \times 720$ units, which needs to be added the facility of an escalator in the path as shown by the dotted line in the figure. Find
(i) The minimum total length of the escalator,
(ii) The heights at which the escalator changes its direction,
(iii) The slopes of the escalator at the turning points.

Solution:
(i) the minimum total length of the escalator.
Shape of the hall in the shopping mall is cuboid. When you open out the cuboid, the not of the cuboid will be as shown in the following diagram.

The path of the escalator is from $O A$ to $A B$ to $B C$ to $C D$
$
\begin{aligned}
& \text { In } \mathrm{OAE}, \mathrm{OA}^2=\mathrm{AE}^2+\mathrm{OE}^2 \\
& \Rightarrow \quad \mathrm{OA}^2=\left[\frac{1}{4}(720)\right]^2+\mathrm{OE}^2 \\
& \Rightarrow \quad \mathrm{OA}^2=(180)^2+(800)^2 \\
& =32400+640000=672400 \\
& \mathrm{OA}=820 \mathrm{~m} \\
& \text { Total length of the escalator }=\mathrm{OA}+\mathrm{AB}+\mathrm{BC}+\mathrm{CD} \\
& =4 \times \mathrm{OA} \quad\left(\text { since } \Delta \mathrm{OAE}=\Delta \mathrm{ABB}^{\prime}=\Delta \mathrm{CDD}^{\prime}\right. \\
& =4 \times 820 \\
&
\end{aligned}
$
The minimum length $=3280$ units
(ii) The height at which the escalator changes its direction.
$
\begin{aligned}
& \mathrm{AE}=\frac{1}{4}(720)=180 \text { units } \\
& \mathrm{BE}=\frac{1}{2}(720)=360 \text { units and } \mathrm{GE}=\frac{3}{4}(720)=540 \text { units }
\end{aligned}
$
(iii) Slope of the escalator at the turning points
Let $\triangle \mathrm{AOE}=\theta$
In $\triangle \mathrm{OAE}, \tan \theta=\frac{\text { opp }}{\mathrm{adj}}=\frac{\mathrm{AE}}{\mathrm{OE}}=\frac{180}{800}=\frac{9}{40}$
$\therefore$ Slope at the point $A=\frac{9}{40}$
Since $\triangle \mathrm{OAE}=\triangle \mathrm{ABB}^{\prime}=\Delta \mathrm{BCC}^{\prime}=\Delta \mathrm{CAD}^{\prime}$
Slope at the points $B, C$ will be $\frac{9}{40}$