Sub-Mathematics, Class-Ix Chapter - Linear Equation in Two ... - Free Printable

Problem: Solve the given worksheet questions on linear equations in two variables.

#### Q.1: Express the following linear equations in the form \( ax + by + c = 0 \) and indicate the values of \( a \), \( b \), and \( c \) in each case:

(i) \( x - \frac{y}{5} - 10 = 0 \)

- The equation is already in the form \( ax + by + c = 0 \).
- Here, \( a = 1 \), \( b = -\frac{1}{5} \), and \( c = -10 \).

(ii) \( y - 2 = 0 \)

- Rewrite the equation as \( 0x + y - 2 = 0 \).
- Here, \( a = 0 \), \( b = 1 \), and \( c = -2 \).

---

#### Q.2: Draw the graph of each of the following linear equations in two variables:

(i) \( y = 3x \)

- This is a linear equation in slope-intercept form \( y = mx + c \), where \( m = 3 \) and \( c = 0 \).
- To draw the graph:
- When \( x = 0 \), \( y = 3(0) = 0 \). So, the point is \( (0, 0) \).
- When \( x = 1 \), \( y = 3(1) = 3 \). So, the point is \( (1, 3) \).
- Plot these points and draw a straight line through them.

(ii) \( 3 = 2x + y \)

- Rewrite the equation as \( 2x + y - 3 = 0 \).
- To draw the graph:
- When \( x = 0 \), \( y = 3 \). So, the point is \( (0, 3) \).
- When \( y = 0 \), \( 2x = 3 \Rightarrow x = \frac{3}{2} \). So, the point is \( \left( \frac{3}{2}, 0 \right) \).
- Plot these points and draw a straight line through them.

---

#### Q.3: If the point \( (3, 4) \) lies on the graph of the equation \( 3y = ax + 7 \), find the value of \( a \).

- Substitute \( x = 3 \) and \( y = 4 \) into the equation \( 3y = ax + 7 \):
\[
3(4) = a(3) + 7
\]
\[
12 = 3a + 7
\]
\[
3a = 12 - 7
\]
\[
3a = 5
\]
\[
a = \frac{5}{3}
\]

---

#### Q.4: Show that the points \( A(1, 2) \), \( B(-1, -16) \), and \( C(0, -7) \) lie on the graph of the linear equation \( y = 9x - 7 \).

- For point \( A(1, 2) \):
\[
y = 9x - 7
\]
Substitute \( x = 1 \) and \( y = 2 \):
\[
2 = 9(1) - 7
\]
\[
2 = 9 - 7
\]
\[
2 = 2 \quad \text{(True)}
\]

- For point \( B(-1, -16) \):
Substitute \( x = -1 \) and \( y = -16 \):
\[
-16 = 9(-1) - 7
\]
\[
-16 = -9 - 7
\]
\[
-16 = -16 \quad \text{(True)}
\]

- For point \( C(0, -7) \):
Substitute \( x = 0 \) and \( y = -7 \):
\[
-7 = 9(0) - 7
\]
\[
-7 = -7 \quad \text{(True)}
\]

Since all three points satisfy the equation \( y = 9x - 7 \), they lie on the graph.

---

#### Q.5: Find \( m \), if the point \( (7, 3) \) lies on the equation \( y - \frac{3}{7} = m(x - \frac{2}{7}) \).

- Substitute \( x = 7 \) and \( y = 3 \) into the equation:
\[
3 - \frac{3}{7} = m \left( 7 - \frac{2}{7} \right)
\]
Simplify both sides:
\[
3 - \frac{3}{7} = \frac{21}{7} - \frac{3}{7} = \frac{18}{7}
\]
\[
7 - \frac{2}{7} = \frac{49}{7} - \frac{2}{7} = \frac{47}{7}
\]
So the equation becomes:
\[
\frac{18}{7} = m \left( \frac{47}{7} \right)
\]
\[
\frac{18}{7} = \frac{47m}{7}
\]
\[
18 = 47m
\]
\[
m = \frac{18}{47}
\]

---

#### Q.6: A fraction becomes \( \frac{1}{3} \) if 2 is added to both numerator and denominator. If 3 is added to both numerator and denominator, it becomes \( \frac{2}{5} \). Assuming the original fraction to be \( \frac{x}{y} \), form a pair of linear equations in two variables for the problem.

- Let the original fraction be \( \frac{x}{y} \).

- Condition 1: If 2 is added to both numerator and denominator, the fraction becomes \( \frac{1}{3} \):
\[
\frac{x + 2}{y + 2} = \frac{1}{3}
\]
Cross-multiply:
\[
3(x + 2) = y + 2
\]
\[
3x + 6 = y + 2
\]
\[
3x - y + 4 = 0
\]

- Condition 2: If 3 is added to both numerator and denominator, the fraction becomes \( \frac{2}{5} \):
\[
\frac{x + 3}{y + 3} = \frac{2}{5}
\]
Cross-multiply:
\[
5(x + 3) = 2(y + 3)
\]
\[
5x + 15 = 2y + 6
\]
\[
5x - 2y + 9 = 0
\]

The pair of linear equations is:
\[
3x - y + 4 = 0
\]
\[
5x - 2y + 9 = 0
\]

---

#### Q.7: Draw the graph of the linear equation \( y = mx + c \) for \( m = \frac{1}{2} \) and \( c = \frac{3}{2} \). Read from the graph the value of \( x \) when \( y = 4.5 \).

- The equation is \( y = \frac{1}{2}x + \frac{3}{2} \).
- To draw the graph:
- When \( x = 0 \), \( y = \frac{1}{2}(0) + \frac{3}{2} = \frac{3}{2} \). So, the point is \( (0, \frac{3}{2}) \).
- When \( y = 0 \), \( 0 = \frac{1}{2}x + \frac{3}{2} \Rightarrow \frac{1}{2}x = -\frac{3}{2} \Rightarrow x = -3 \). So, the point is \( (-3, 0) \).
- Plot these points and draw a straight line through them.

- To find \( x \) when \( y = 4.5 \):
\[
4.5 = \frac{1}{2}x + \frac{3}{2}
\]
\[
4.5 = \frac{1}{2}x + 1.5
\]
\[
4.5 - 1.5 = \frac{1}{2}x
\]
\[
3 = \frac{1}{2}x
\]
\[
x = 6
\]

---

#### Q.8: Draw the graph of the linear equation \( 3x + 4y = 6 \). At what points does the graph cut the \( x \)-axis and \( y \)-axis?

- To find the \( x \)-intercept (where \( y = 0 \)):
\[
3x + 4(0) = 6
\]
\[
3x = 6
\]
\[
x = 2
\]
So, the \( x \)-intercept is \( (2, 0) \).

- To find the \( y \)-intercept (where \( x = 0 \)):
\[
3(0) + 4y = 6
\]
\[
4y = 6
\]
\[
y = \frac{3}{2}
\]
So, the \( y \)-intercept is \( \left( 0, \frac{3}{2} \right) \).

- Plot the points \( (2, 0) \) and \( \left( 0, \frac{3}{2} \right) \) and draw a straight line through them.

---

#### Q.9: If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is 2 units and 0 units.

- Let the work done be \( W \) and the distance travelled be \( d \). Since \( W \) is directly proportional to \( d \), we have:
\[
W = kd
\]
where \( k \) is the constant of proportionality. Given \( k = 5 \):
\[
W = 5d
\]

- To draw the graph:
- When \( d = 0 \), \( W = 5(0) = 0 \). So, the point is \( (0, 0) \).
- When \( d = 1 \), \( W = 5(1) = 5 \). So, the point is \( (1, 5) \).
- Plot these points and draw a straight line through them.

- From the graph:
- When \( d = 2 \), \( W = 5(2) = 10 \).
- When \( d = 0 \), \( W = 5(0) = 0 \).

---

#### Q.10: The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as \( x \) km and total fare as \( Rs \, y \), write a linear equation for this information, and draw its graph.

- For the first kilometre, the fare is Rs 8.
- For the subsequent distance \( (x - 1) \) km, the fare is \( 5(x - 1) \) Rs.
- Total fare \( y \) is:
\[
y = 8 + 5(x - 1)
\]
Simplify:
\[
y = 8 + 5x - 5
\]
\[
y = 5x + 3
\]

- To draw the graph:
- When \( x = 0 \), \( y = 5(0) + 3 = 3 \). So, the point is \( (0, 3) \).
- When \( x = 1 \), \( y = 5(1) + 3 = 8 \). So, the point is \( (1, 8) \).
- Plot these points and draw a straight line through them.

---

Final Answers:

\[
\boxed{
\begin{aligned}
&\text{Q.1: (i) } a = 1, b = -\frac{1}{5}, c = -10; \quad (ii) a = 0, b = 1, c = -2 \\
&\text{Q.3: } a = \frac{5}{3} \\
&\text{Q.4: Points lie on the graph.} \\
&\text{Q.5: } m = \frac{18}{47} \\
&\text{Q.6: } 3x - y + 4 = 0, \, 5x - 2y + 9 = 0 \\
&\text{Q.7: } x = 6 \\
&\text{Q.8: } x\text{-intercept } (2, 0), \, y\text{-intercept } \left( 0, \frac{3}{2} \right) \\
&\text{Q.9: } W = 5d; \, W = 10 \text{ when } d = 2, \, W = 0 \text{ when } d = 0 \\
&\text{Q.10: } y = 5x + 3
\end{aligned}
}
\]