Exercise 6.3

1. Draw the graph of each of the following linear equations.

(i) 2y = −x + 1

Solution:

Let's find two points:

When x = 0, 2y = 1 => y = 122. So, (, 122) is a point.

When x = 1, 2y = -1 + 1 => 2y = => y = . So, (, ) is a point.

(ii) –x + y = 6

Solution:

Let's find two points:

When x = 0, y = . So, (, ) is a point.

When y = 0, -x = => x = . So, (, ) is a point.

(iii) 3x + 5y = 15

Solution:

Let's find two points:

When x = 0, 5y = 15 => y = . So, (, ) is a point.

When y = 0, 3x = 15 => x = . So, (, ) is a point.

(iv) x2 − y3 = 2

Solution:

Let's find two points:

When x = 0, -y/3 = 2 => y = . So, (, ) is a point.

When y = 0, x/2 = 2 => x = . So, (, ) is a point.

2. Draw the graph of each of the following linear equations and answer the following question.

(i) y = x (ii) y = 2x (iii) y = −2x (iv) y = 3x (v) y = −3x

Solution:

Lets form the tables for drawing the graphs,

(i) y = x

(ii) y = 2x

(iii) y = -2x

(iv) y = 3x

(v) y = -3x

(i) Are all these equations of the form y = mx, where m is a real number?

Solution:

Yes, all these equations are of the form y = mx, where m is a real number (the coefficient of x).

(ii) Are all these graphs passing through the origin?

Solution:

Yes, if you substitute x = 0 in any of these equations, you get y = 0. This means all the lines pass through the point (0, 0), which is the origin.

(iii) What can you conclude about these graphs?

Solution:

We can conclude that equations of the form y = mx represent straight lines passing through the origin. The value of 'm' affects the slope (steepness) of the line.

3. Draw the graph of the equation 2x + 3y = 11. Find the value of y when x = 1 from the graph.

Solution:

The values for drawing the graph.

4. Draw the graph of the equation y − x = 2. Find from the graph

(i) the value of y when x = 4

(ii) the value of x when y =-3

Solution:

The values for drawing the graph.

5. Draw the graph of the equation 2x+3y=12. Find the solutions from the graph

(i) Whose y-coordinate is 3

(ii) Whose x-coordinate is -3

Solution:

6. Draw the graph of each of the equations given below and also find the coordinates of the points where the graph cuts the coordinate axes

(i) 6x − 3y = 12

Solution:

7. Rajiya and Preethi, two students of Class IX, together collected ₹1000 for the Prime Minister Relief Fund for victims of natural calamities. Write a linear equation and draw a graph to depict the statement.

Solution:

Let the amount collected by Rajiya be ₹x and the amount collected by Preethi be ₹y.

According to the problem, x + y =

To draw the graph, we can find two points that satisfy this equation:

When x = 0, y = . So, (, ) is a point on the line.

When y = 0, x = . So, (, ) is another point on the line.

8. Gopaiah sowed wheat and paddy in two fields of total area 5000 square meters. Write a linear equation and draw a graph to represent the same.

Solution:

Let the area of the field where wheat is sowed be x square meters and the area of the field where paddy is sowed be y square meters.

According to the problem, x + y =

To draw the graph, we can find two points that satisfy this equation:

When x = 0, y = . So, (, ) is a point on the line.

When y = 0, x = . So, (, ) is another point on the line.

9. The force applied on a body of mass 6 kg is directly proportional to the acceleration produced in the body. Write an equation to express this observation and draw the graph of the equation.

Solution:

Let the force applied be F and the acceleration produced be a.

Since F is directly proportional to a, we can write F = kaa, where k is the constantVariable of proportionality.

We know that F = maa

Where m is the mass of the body.

Therefore, ma = ka

Since m = kg.

6a6ma = ka

Comparing both sides, we get k = .

So, the equation becomes F = 6aka.

To draw the graph, we can find two points that satisfy this equation:

When a = 0, F = .

So, (, ) is a point on the line.

When a = 1, F = .

So, (, ) is another point on the line.

10. A stone is falling from a mountain. The velocity of the stone is given by V = 9.8t. Draw its graph and find the velocity of the stone 4 seconds after start.

Solution:

The equation V = 9.8t represents a linear equation where V is the velocityVolume and t is the .

To draw the graph, we can find two points that satisfy this equation:

When t = 0, V = .

So, (, ) is a point on the line.

When t = 1, V = .

So, (, ) is another point on the line.

To find the velocity of the stone 4 seconds after the start, substitute t = 4 in the equation V = 9.8t.

V = 9.8 × 4 = m/s.

Therefore, the velocity of the stone 4 seconds after the start is 39.2 m/s.

Now, plot these two points on a graph paper and draw a straight line passing through them. This line represents the equation V = 9.8t.