Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.9 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.9

Question 1.
Fill in the blanks:
(i) y = p x where p ∈ Z always passes through the _________ .
Answer:
Origin (0,0)
Hint:
[When we substitute x = 0 in equation, y also becomes zero. (0,0) is a solution]

Samacheer Kalvi 8th Maths Guide Chapter 3 Algebra Ex 3.9

(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 cm = _________ units
y-axis 1 cm = _________ units
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 1
Answer:
3 units, 25 units
Hint:
With reference to given graph,
On the x – axis. 1 cm = 3 units
y axis, 1 cm = 25 units

Samacheer Kalvi 8th Maths Guide Chapter 3 Algebra Ex 3.9

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) y = -9x not passes through the origin.
Answer:
False
Hint:
y = -9x substituting for x as zero, we get y = -9 × 0 = 0
∴ for x = 0, 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
ax + by = 1 Where a & b are the x & y intercept respectively.
Since the intercept points are (2. 0) & (0, 2)
a = 2, b = 2
∴ 2x + 2y = 1
When the point (2. 2) is considered & substituted in the equation
2x + 2y = 1, we get
2 × 2 + 2 × 2 = 4 ≠ 1
∴ the point (2. 2) does not satisfy the equation. Therefore the line does not pass through (2, 2)
Alternatively graphical method
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 2
as we can see the line doesn’t pass through (2, 2)

Samacheer Kalvi 8th Maths Guide Chapter 3 Algebra Ex 3.9

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)
y – axis intercept point – (0, 2) using 2 point formula.
amacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 3
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 4
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 5
Any point in II quadrant will have x as negative & y as positive.
So let us take x value as – 2
∴ -2 + y = 2
∴ y = 2 + 2 = 4
∴ Point in II Quadrant is (-2, 4)

Samacheer Kalvi 8th Maths Guide Chapter 3 Algebra Ex 3.9

Question 5.
If the points P(5, 3) Q(-3, 3) R (-3, -4) and S form a rectangle then find the coordinate of 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 PQ & QR should be paraHel to PS
  • S should lie on the straight line from R parallel to x-axis & straight line from P parallel to y-axis
  • Therefore, we get S to be (5, -4)

[Note: We don’t need graph sheet for approximate plotting. This is just for graphical understanding]
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 6

Samacheer Kalvi 8th Maths Guide Chapter 3 Algebra Ex 3.9

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]
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 7

  • 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)

Samacheer Kalvi 8th Maths Guide Chapter 3 Algebra Ex 3.9

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:
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 8
Equation of line joining 2 points by 2 point formula is given by
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 9
Cross multiplying, we get
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 10
Transposing the variables, we get
11 x + 5 y = 35 – 33 = 2
11 x + 5y = 2 – Line 1
Similarly, we should find out equation of second line
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 11
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 12
∴ 9y – 54 = 13x – 52
∴ 9y – 13x = 2 – Line 2
For finding point of intersection, we need to solve the 2 line equation to find a point that will satisfy both the line equations.
∴ Solving for x & y from line 1 & line 2 as below
11x + 5y = 2 ⇒ multiply both sides by 13,
11 × 13x + 5 × 13y = 26 …….. (3)
Line 2: 9y – 13x = 2 ⇒ multiply both sides by 11
9 × 11y – 13 × 11x = 22 ……… (4)
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 13
∴ 164 y = 48
∴ y = \(\frac{48}{164}=\frac{12}{41}\)
Substituting this value ofy in line I we get
11 x + 5 y = 2
11 x + 5 × \(\frac{12}{41}\) = 2
11 x = 2 – \(\frac{60}{41}=\frac{82-60}{41}=\frac{22}{41}\)
∴ x = \(\frac{2}{41}\)
[∴ Point of intersection is \(\left(\frac{2}{41}, \frac{12}{41}\right)\)]
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’
∴ put y = 0 in above equation
∴ 11 x – 5 × 0 = 2
∴ 11x + 0 = 2
∴ x = \(\frac{2}{11}\)
∴ [Point is \(\left(\frac{2}{11}, 0\right)\)]
Similarly, Point of intersection of line with y – axis is when x-coordinate becomes ‘0’
∴ put x = 0 in above equation
∴ 11 × 0 + 5y = 2
∴ 0 + 5y = 2
y = \(\frac{2}{5}\)
∴ [Point is \(\left(0, \frac{2}{5}\right)\)]
Similarly for line 2,
9y – 13x = 2
For finding x intercept, i.e point where line meets x axis, we know that y coordinate becomes ‘0’
∴ Substituting y = 0 in above eqn. we get
9 × 0 – 13x = 2
∴ 0 – 13x = 2
∴ x = \(\frac{-2}{13}\)
∴ [Point: \(\left(\frac{-2}{13}, 0\right)\)]
Similarly for y – intercept, x – coordinate becomes ‘0’,
∴ Substituting for x = 0 in above equation, we get
9 y – 13 × 0 = 2
9y – 0 = 2
9y = 2
y = \(\frac{2}{9}\)
[Point \(\left(0, \frac{2}{9}\right)\)]

Samacheer Kalvi 8th Maths Guide Chapter 3 Algebra Ex 3.9

Question 8.
Draw the graph of the following equations: (i) x = – 7 (ii) y = 6
Answer:
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 14

Question 9.
Draw the graph of
(i) y = – 3x
(ii ) y = x – 4
(ii) y = 2x + 5
Answer:
To draw graph, we need to find out some points.
(i) y = – 3x
for y = -3x, let us first substituting values & check
put x = 0
y = 3 × 0 = 0
∴ (0,0) is a point
put x = 1
y = -3 × 1 = – 3
∴ (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
y = 0 – 4 = – 4
∴ (0, – 4) is a point
x = 4
y = 4 – 4 = 0
∴ (4, 0) is a point

(iii) y = 2x + 5
for y = 2x + 5
put x = – 1
y = 2(-1) + 5 = – 2 + 5 = 3
∴ (-1, 3) is a point
put x = – 2
y = 2(-2) + 5 = – 4 + 5 = 1
∴ (-2, 1) is a point
Now let us plot the points & join them on graph
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 15

Samacheer Kalvi 8th Maths Guide Chapter 3 Algebra Ex 3.9

Question 10.
Find the values
(a) y = x + 3
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 16
Answer:
Let y = x+3
(i) if x = 0, y = 0 + 3 = 3,
∴ y = 3
(ii) y = 0, 0 = x + 3,
∴ x = – 3
(iii) x = -2, y = -2 + 3,
∴ y = 1
(iv) y = – 3, – 3 = x + 3,
∴ x = – 6

y = x + 3
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 17

Samacheer Kalvi 8th Maths Guide Chapter 3 Algebra Ex 3.9

(b) 2x + y – 6 = 0
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 18
Answer:
Let 2x + y – 6 = 0
(i) x = 0 2 × 0 + y – 6 = 0 ∴ y = 6
(ii) y = 0, 2x + 0 – 6 = 0, ∴ 2x = 6
x = 3 ,
(iii) x = -1, 2 × (-1) + y – 6 = 0, 8 + y = 0
y = 8
(iv) y = -2, 2x – 2 – 6 = 0, 2x = 8
x = 4

2x + y – 6 = 0
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 19

Samacheer Kalvi 8th Maths Guide Chapter 3 Algebra Ex 3.9

(c) y = 3x + 1
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 20
Answer:
(i) x = -1, y = 3(-1) + 1 = 0
∴ y = -2
(ii) x = 0, y = 3(0) + 1 = 0
∴ y = 1
(iii) x = 1, y = 3(1) + 1 = 0
∴ y = 4
(iv) x = 2, y = 3(2) + 1 = 0
∴ y = 7

y = 3x + 1
Samacheer Kalvi 8th Maths Guide Answers Chapter 3 Algebra Ex 3.9 21