Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.4
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.4
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.4
Question 1.
Fill in the blanks:
Question (i)
y = px 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]
Question (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)
Question (iii)
Scale for the given graph, on the x – axis 1 cm = ……… units y – axis 1 cm = ………. units
Answer:
3 units, 25 units
Hint:
With reference to given graph,
On the x – axis, 1cm = 3 units
y axis, 1cm = 25 units
Question 2.
Say True or False:
Question (i)
The points (1, 1) (2, 2) (3, 3) lie on a 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
Question (ii)
y = – 9x not passes through the origin.
Answer:
False
Hint:
y = – 9x substituting forx as zero, we get y = – 9 x 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).
Solution:
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 x 2 + 2 x 2 = 4 ≠ 1
∴ The point (2, 2) does not satisfy the equation. Therefore the line does not pass through (2, 2)
Alternatively graphical method:
as we can see the line doesn’t 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.
Solution:
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.
Solution:
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 parallel 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]
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?
Solution:
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.
Solution:
∴ 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 x 13x + 5 x 13y = 26
Line 2:
9y – 13x = 2 ⇒ multiply both sides by 11
9 x 11y – 13 x 11x = 22
Solution:
Question 9.
Draw the graph of
- y = – 3x
- y = x – 4
Solution:
To draw graph, we need to find out some points.
1. for y = – 3x, let us first substituting values & check
put x = 0
y = – 3 x 0 = 0 ∴ (0,0) is a point
put x = 1
y = – 3 x 1 = – 3 ∴ (1, – 3) is a point
If join these 2 points, we will get the line
2. 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
Now let us plot the points & join them on graph
Question 10.
Find the values
Solution:
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
Let 2x + 7 – 6 = 0
(i) x = 0
2 x 0 + y – 6 = 0
∴ 7 = 6
(ii) y = 0
2x + 0 – 6 = 0
2x = 6
x = 3
(iii) x = – 2
2 x (- 2) + y – 6 = 0
y – 10 = 0
y = 10
(iv) y = -3
2x – 3 – 6 = 0
2x = 9
x = 9/2
Steps :
- Draw the axis, x – axis = radius & y – axis = circumference
- Mark the points by choosing scale of 1 cm = 7 units for x axis & 1 cm = 44 units for y – axis
- Join the points to form a line.
- Now for r value = 24.5 (mid value between 21 & 28), draw vertical line to touch the main line & from there drop line parallel to x axis and note where it meets y axis.
It meets between 132 & 176. Taking the mid value
i.e 132+176/2 = 154, we get circumference of 154 when r = 24.5.
Similarly, for circumference of 242, we get r value to be 38.5
Question 12.
An over-head tank is full with water. Water leaks out from it, at a constant rate of 10 litres per hour. Draw a “time-wastage” graph for this situation and find
- The water wasted in 150 minutes
- The time at which 75 litres of water is wasted.
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