Exercise 3.9 - Chapter 3 Algebra 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 1.
Fill in the blanks:
(i) $y=p x$ where $p \in Z$ always passes through the
Answer:
Origin $(0,0)$
Hint:
[When we substitute $\mathrm{x}=0$ in equation, $\mathrm{y}$ also becomes zero. $(0,0)$ is a solution]
(ii) The intersecting point of the line $x=4$ and $y=-4$ is
Answer:
$4,-4$
Hint:
$x=4$ is a line parallel to the $y$ - axis and
$y=-4$ is a line parallel to the $x$-axis. The point of intersection is a point that lies on both lines \& which should satisfy both the equations. Therefore, that point is $(4,-4)$
(iii) Scale for the given graph,
On the $x$-axis $1 \mathrm{~cm}=$ __units
$\mathrm{y}$-axis $1 \mathrm{~cm}=$ __units

Answer:
3 units, 25 units
Hint:
With reference to given graph,
On the $x$-axis. $1 \mathrm{~cm}=3$ units
y axis, $1 \mathrm{~cm}=25$ units
 

Question $2 .$
Say True or False.
(i) The points ( 1,1$)(2,2)(3,3)$ lie on a same straight line.
Answer:
True
Hint:
The points $(1,1),(2,2),(3,3)$ all satisfy the equation $y=x$ which is straight line.
Hence, it is true
(ii) $\mathrm{y}=-9 \mathrm{x}$ not passes through the origin.
Answer:
False
Hint:
$\mathrm{y}=-9 \mathrm{x}$ substituting for $\mathrm{x}$ as zero, we get $\mathrm{y}=-9 \times 0=0$
$\therefore$ for $\mathrm{x}=0, \mathrm{y}=0$. Which means line passes through $(0,0)$, hence statement is false.
 

Question $3 .$
Will a line pass through $(2,2)$ if it intersects the axes at $(2,0)$ and $(0,2)$.
Answer:

Given a line intersects the axis at $(2,0) \&(0,2)$
Let line intercept form be expressed as
$\mathrm{ax}+\mathrm{by}=1$ Where $\mathrm{a} \& \mathrm{~b}$ are the $\mathrm{x} \& \mathrm{y}$ intercept respectively.
Since the intercept points are $(2.0) \&(0,2)$
$\begin{aligned}
&\mathrm{a}=2, \mathrm{~b}=2 \\
&\therefore 2 \mathrm{x}+2 \mathrm{y}=1
\end{aligned}$
When the point $(2.2)$ is considered \& substituted in the equation
$2 \mathrm{x}+2 \mathrm{y}=1 \text {, we get }$
$2 \times 2+2 \times 2=4 \neq 1$
$\therefore$ the point $(2.2)$ does not satisfy the equation. Therefore the line does not pass through $(2,2)$

Question $4 .$
A line passing through $(4,-2)$ and intersects the $Y$-axis at $(0,2)$. Find a point on the line in the second quadrant.
Answer:
Line passes through $(4,-2)$
$\mathrm{y}$ - axis intercept point $-(0,2)$ using 2 point formula.

$$
\begin{aligned}
\frac{y-y_{1}}{x-x_{1}} &=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\
\frac{y-2}{x-0} &=\frac{-2-2}{4-0} \\
\therefore \frac{y-2}{x} &=\frac{-4}{4}=-1 \\
y-2 &=-1 \times x \\
\therefore y-2 &=-x \\
\therefore x+y &=2 \text { is the equation of the line. }
\end{aligned}
$$
Any point in II quadrant will have $\mathrm{x}$ as negative \& $\mathrm{y}$ as positive.
So let us take $x$ value as $-2$
$$
\begin{aligned}
&\therefore-2+\mathrm{y}=2 \\
&\therefore \mathrm{y}=2+2=4
\end{aligned}
$$
$\therefore$ Point in II Quadrant is $(-2,4)$
 

Question $5 .$
If the points $\mathrm{P}(5,3) \mathrm{Q}(-3,3) \mathrm{R}(-3,-4)$ and $\mathrm{S}$ form a rectangle then find the coordinate of $\mathrm{S}$.
Answer:
Plotting the points on a graph (approximately)
Steps:
- Plot P, Q, R approximately on a graph.
- As it is a rectangle, RS should be parallel to $\mathrm{PQ} \& \mathrm{QR}$ should be paraHel to $\mathrm{PS}$
- S should lie on the straight line from $R$ parallel to $\mathrm{x}$ axis \& straight line from $P$ parallel to $\mathrm{y}$-axis
- Therefore, we get $S$ to be $(5,-4)$

Question $6 .$
A line passes through $(6,0)$ and $(0,6)$ and an another line passes through $(-3,0)$ and $(0,-3)$. What are the points to be joined to get a trapezium?
Answer:
In a trapezium. there are 2 opposite sides that are parallel. The other opposite sides are non-parallel.
Now, let us approximately plot the points for our understanding
[no need of graph sheet]
- Plot the points $(0,6),(6,0),(-3,0) \&(0,-3)$
- Join $(0,6) \&(6,0)$
- Join $(-3,0) \&(0,-3)$
- We find that the lines formed by joining the points are parallel lines.
- So, for forming a trapezium, we should join $(0,6),(-3,0) \&(0,-3),(6,0)$
 

Question $7 .$
Find the point of intersection of the line joining points $(-3,7)(2,-4)$ and $(4,6)(-5$, $-7)$.Also find the point of intersection of these lines and also their intersection with the axis.
Answer:
Line 1: Joining points $(-3,7) \&(2,-4)$
Equation of line joining 2 points by 2 point formula is given by
$\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$\therefore \frac{y-7}{x-(-3)}=\frac{-4-7}{2-(-3)}$
$\begin{aligned}
\frac{y-7}{x+3} &=\frac{-11}{2+3} \\
\therefore \frac{y-7}{x+3} &=\frac{-11}{5}
\end{aligned}$
Cross multiplying, we get
$\begin{aligned}
&\frac{y-7}{x+3}=\frac{-11}{5} \\
&5(y-7)=-11(x+3) \\
&5 y-35=-11 x-33
\end{aligned}$
Transposing the variables, we get
$11 x+5 y=35-33=2$
$11 x+5 y=2$ - Line 1
Similarly, we should find out equation of second line

Joining the points $(4,6) \&(-5,-7)$
$\begin{aligned}
& \frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\
\therefore & \frac{y-6}{x-4}=\frac{-7-6}{-5-4} \\
\therefore & \frac{y-6}{x-4}=\frac{-13}{-9}=\frac{13}{9}
\end{aligned}$
$\begin{aligned}
&\therefore 9 \mathrm{y}-54=13 \mathrm{x}-52 \\
&\therefore 9 \mathrm{y}-13 \mathrm{x}=2 \text { - Line } 2
\end{aligned}$
For finding point of intersection, we need to solve the 2 line equation to find a point that will satisfy both the line equations.
$\therefore$ Solving for $x$ \& $y$ from line 1 \& line 2 as below
$11 x+5 y=2 \Rightarrow$ multiply both sides by 13
$11 \times 13 \mathrm{x}+5 \times 13 \mathrm{y}=26 \ldots \ldots . .(3)$
Line $2: 9 y-13 x=2 \Rightarrow$ multiply both sides by 11
$9 \times 11 y-13 \times 11 x=22$ (4)

Adding 3 & 4

$\therefore 164 \mathrm{y}=48$
$\therefore \mathrm{y}=\frac{48}{164}=\frac{12}{41}$
Substituting this value ofy in line I we get
$11 \mathrm{x}+5 \mathrm{y}=2$
$11 x+5 \times \frac{12}{41}=2$
$11 \mathrm{x}=2-\frac{60}{41}=\frac{82-60}{41}=\frac{22}{41}$
$\therefore \mathrm{x}=\frac{2}{41}$
$\left[\therefore\right.$ Point of intersection is $\left.\left(\frac{2}{41}, \frac{12}{41}\right)\right]$
To find point of intersection of the lines with the axis, we should substitute values \& check

Line 1: $11 x+5 y=2$
Point of intersection of line with $x$ - axis, i.e $y$ coordinate is ' 0 '
$\therefore$ put $\mathrm{y}=0$ in above equation
$\therefore 11 \mathrm{x}-5 \times 0=2$
$\therefore 11 \mathrm{x}+0=2$
$\therefore x=\frac{2}{11}$
$\therefore$ [Point is $\left(\frac{2}{11}, 0\right)$ ]
Similarly, Point of intersection of line with $y$-axis is when $x$-coordinate becomes ' 0 '
$\therefore$ put $\mathrm{x}=0$ in above equation
$\therefore 11 \times 0+5 \mathrm{y}=2$
$\therefore 0+5 y=2$
$\mathrm{y}=\frac{2}{5}$
$\therefore$ [Point is $\left(0, \frac{2}{5}\right)$ ]
Similarly for line 2 ,
$9 y-13 x=2$
For finding $x$ intercept, i.e point where line meets $x$ axis, we know that $y$ coordinate becomes ' 0 '
$\therefore$ Substituting $y=0$ in above eqn. we get
$9 \times 0-13 x=2$

$\therefore 0-13 \mathrm{x}=2$ $\therefore \mathrm{x}=\frac{-2}{13}$ $\therefore\left[\right.$ Point: $\left.\left(\frac{-2}{13}, 0\right)\right]$
Similarly for $\mathrm{y}$ - intercept, $\mathrm{x}$ - coordinate becomes ' 0 ',
$\therefore$ Substituting for $\mathrm{x}=0$ in above equation, we get
$9 y-13 \times 0=2$
$9 y-0=2$
$9 y=2$
$\mathrm{y}=\frac{2}{9}$
[Point $\left(0, \frac{2}{9}\right)$ ]

Question $8 .$
Draw the graph of the following equations: (i) $x=-7$ (ii) $y=6$

Answer:

Question $9 .$
Draw the graph of
(i) $y=-3 x$
(ii) $y=x-4$
(ii) $\mathrm{y}=2 \mathrm{x}+5$
Answer:
To draw graph, we need to find out some points.
(i) $y=-3 x$
for $y=-\underline{3 x}$, let us first subbstituting values \& check
put $x=0$
$\mathrm{y}=3 \times 0=0$
$\therefore(0,0)$ is a point
put $\mathrm{x}=1$
$y--3 \times 1--3$
$\therefore(1,-3)$ is a point
If join these 2 points, we will get the line
(ii) $y=x-4$
for $y=x-4$
put $x=0$
$\mathrm{y}=0-4=-4$
$\therefore(0,-4)$ is a point
$\mathrm{x}=4$

$\mathrm{y}=4-4=0$
$\therefore(4,0)$ is a point
(iii) $y=2 x+5$
for $y=2 x+5$
put $\mathrm{x}=-1$
$y=2(-1)+5=-2+5=3$
$\therefore(-1,3)$ is a point
put $\mathrm{x}=-2$
$y=2(-2)+5=-4+5=1$
$\therefore(-2,1)$ is a point
Now let us plot the points \& join them on graph

Question $10 .$
Find the values
(a) $y=x+3$

Answer:
Let $y=x+3$
(i) if $x=0, y=0+3=3$,
$\therefore \mathrm{y}=3$
(ii) $\mathrm{y}=0,0=\mathrm{x}+3$,
$\therefore x=-3$
(iii) $x=-2, y=-2+3$,
$\therefore \mathrm{y}=1$
(iv) $y=-3,-3=x+3$,
$\therefore \mathrm{x}=-6$

(b) $2 x+y-6=0$
Answer:
Let $2 x+y-6=0$
(i) $x=02 \times 0+y-6=0 \therefore y=6$
(ii) $\mathrm{y}=0,2 \mathrm{x}+0-6=0, \therefore 2 \mathrm{x}=6$ $x=3$,
(iii) $\mathrm{x}=-1,2 \times(-1)+\mathrm{y}-6=0,8+\mathrm{y}=0$ $y=8$
(iv) $\mathrm{y}=-2,2 \mathrm{x}-2-6=0,2 \mathrm{x}=8$ $x=4$

(c) $y=3 x+1$
Answer:
(i) $x=-1, y=3(-1)+1=0$
$\therefore \mathrm{y}=-2$
(ii) $x=0, y=3(0)+1=0$
$\therefore \mathrm{y}=1$
(iii) $x=1, y=3(1)+1=0$
$\therefore \mathrm{y}=4$
(iv) $x=2, y=3(2)+1=0$
$\therefore \mathrm{y}=7$