# Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.4

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## 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 ………
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 ………
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

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.
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.
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:

Line passes through (4, – 2)
y – axis intercept point – (0, 2)
using 2 point formula,
$$\frac { y-y_{ 1 } }{ x-x_{ 1 } }$$ = $$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$$
$$\frac{y-2}{x-0}$$ = $$\frac{1-2-2}{4-0}$$
∴ $$\frac{y-2}{x}$$ = $$\frac{-4}{4}$$ = – 1
y – 2 = – 1 × x

∴ x + y = 2 is the equation of the line.
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)

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:

1. Plot P, Q, R approximately on a graph.
2. As it is a rectangle, RS should be parallel to PQ & QR should be parallel to PS
3. S should lie on the straight line from R parallel to x – axis & straight line from P parallel to y – axis
4. 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]

1. Plot the points (0, 6), (6, 0), (-3, 0) & (0, – 3)
2. Join (0, 6) & (6, 0)
3. Join (-3, 0) & (0, -3)
4. We find that the lines formed by joining the points are parallel lines.
5. 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:

Equation of line joining 2 points by 2 point formula is given by
$$\frac { y-y_{ 1 } }{ x-x_{ 1 } }$$ = $$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$$
∴ $$\frac{y-7}{x-(-3)}$$ = $$\frac{-4-7}{2-(-3)}$$
$$\frac{y-7}{x+3}$$ = $$\frac{-11}{2+3}$$
∴ $$\frac{y-7}{x+3}$$ = $$\frac{-11}{5}$$
Cross multiplying, we get

Transposing the variables, we get
11x + 5y = 35 – 33 = 2
11x + 5y = 2 – Line 1
Similarly, we should find out equation of second line

∴ 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

Substituting this value of y in line 1 we get
11x + 5y = 2
11x + 5 x $$\frac{12}{41}$$ = 2
11x = 2 – $$\frac{60}{41}$$ = $$\frac{82-60}{41}$$ = $$\frac{22}{41}$$
x = $$\frac{2}{41}$$
∴ Point of intersection is ($$\frac{2}{41}$$ $$\frac{12}{41}$$)
To find point of intersection of the lines with the axis, we should substitute values & check
Line 1:
11x + 5y = 2
Point of intersection of line with x – axis, i.e y coordinate is ‘0’
∴ put y = 0 in above equation
∴ 11x – 5 x 0 = 2
11x + 0 = 2
x = $$\frac{2}{11}$$
∴ Point is ($$\frac{2}{11}$$, 0)
Similarly, point of intersection of line with y – axis is when x – coordinate becomes ‘0’
∴ put x = 0 in above equation
∴ 11 x 0 + 5y = 2
∴ 0 + 5y = 2
y = $$\frac{2}{5}$$
∴ Point is (0, $$\frac{2}{5}$$)
Similarly for line 2,
9y – 13x = 2
For finding x intercept, i.e point where line meets x axis, we know thaty coordinate becomes ‘0’
∴ Substituting y – 0 in above eqn. we get
9 x 0 – 13x = 2
∴ 0 – 13x = 2
x = $$\frac{-2}{13}$$
∴ Point is ($$\frac{-2}{13}$$, 0)
Similarly for y – intercept, x – coordinate becomes ‘0’,
∴ Substituting for x = 0 in above equation, we get
9y – 13 x 0 = 2
9y – 0 = 2
9y = 2
y = $$\frac{2}{9}$$
Point is (0, $$\frac{2}{9}$$)

Question 8.
Draw the graph of the following equations:

1. x = – 7
2. y = 6

Solution:

Question 9.
Draw the graph of

1. y = – 3x
2. 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 = $$\frac{9}{2}$$

Question 11.
The following is a table of , values connecting the radii of a few circles and their circumferences (Taking π = $$\frac{22}{7}$$) Illustrate the relation with a graph and find

1. The radius when the circumference is 242 units.
2. The circumference when the radius is 24.5 units.

Solution:

r = 24.5 Circumference?

Steps :

1. Draw the axis, x – axis = radius & y – axis = circumference
2. Mark the points by choosing scale of 1 cm = 7 units for x axis & 1 cm = 44 units for y – axis
3. Join the points to form a line.
4. 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 $$\frac{132+176}{2}$$ = 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

1. The water wasted in 150 minutes
2. The time at which 75 litres of water is wasted.

Solution:
From over head tank, water leaks out at 10 l/hr.
∴ Let us see how much leakage happens with time
Time     Leakage
1 hr        10 ltrs.
2 hrs       20 ltrs.
3 hrs       30 ltrs.
4 hrs       40 ltrs.. and so on
150 min = 120 min + 30 min
= 2 hr 30 min = 2 $$\frac{1}{2}$$ hrs
Now let us plot a graph between time & water leakage

Water wasted in 150 min (2$$\frac{1}{2}$$ hrs) = 25 litres
Time at which 75 litre of water is wasted = 7$$\frac{1}{2}$$ hrs = 450 minutes.