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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 ………

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:

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:

- 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:

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:

- x = – 7
- y = 6

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 = \(\frac{9}{2}\)

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

- The radius when the circumference is 242 units.
- The circumference when the radius is 24.5 units.

Solution:

r = 24.5 Circumference?

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 \(\frac{132+176}{2}\) = 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

- The water wasted in 150 minutes
- 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.