CBSE Class 9 Mathematics Linear Equations In Two Variables Worksheet Set 06

Main Concepts and Results

A mathematical statement expressing the equality of two algebraic quantities is defined as an equation. When an equation is written in the general form \( ax + by + c = 0 \), where the coefficients \( a \), \( b \), and constant \( c \) are real numbers with \( a \neq 0 \) and \( b \neq 0 \), it is called a linear equation in two variables. Determining the values of the variables that satisfy the relationship is termed solving the equation.

The set of solutions of any linear equation remains completely unchanged under the following transformations:

• Adding or subtracting the exact same real value on both sides of the equation.
• Multiplying or dividing both sides of the equation by the same non-zero real quantity.

Any linear equation in two variables possesses an infinite number of distinct solutions. When plotted on the Cartesian plane, the geometric representation of such an equation forms a straight line, and every single coordinate pair lying on this line corresponds to a valid solution. Conversely, each solution pair can be located as a specific coordinate point on the line. Furthermore, graphs of the form \( x = a \) represent vertical lines parallel to the y-axis, whereas graphs of the form \( y = a \) represent horizontal lines parallel to the x-axis.

(B) Multiple Choice Questions

Sample Question 1. The linear equation 3x – y = x – 1 has :
(a) A unique solution
(b) Two solutions
(c) Infinitely many solutions
(d) No solution
Answer: (c) Infinitely many solutions
Subtracting \( x \) from both sides simplifies the equation to \( 2x - y + 1 = 0 \). Because this is a standard linear equation with two variables, selecting any real value for \( x \) yields a corresponding real value for \( y \). Hence, it admits an endless number of solutions.
In simple words: A linear equation with two variables always has an infinite number of coordinate pairs that satisfy it.

Exam Tip: Never confuse a single linear equation in two variables (which always has infinitely many solutions) with a system of two simultaneous linear equations.

Sample Question 2. A linear equation in two variables is of the form ax + by + c = 0, where
(a) a ≠ 0, b ≠ 0
(b) a = 0, b ≠ 0
(c) a ≠ 0, b = 0
(d) a = 0, c = 0
Answer: (a) a ≠ 0, b ≠ 0
For the expression \( ax + by + c = 0 \) to genuinely contain two distinct variables, the coefficients of both \( x \) and \( y \) must be non-zero at the same time. If either coefficient were zero, the equation would reduce to an equation in only one variable.
In simple words: To keep both x and y in the equation, neither a nor b can be zero.

Exam Tip: Remember the condition \( a^2 + b^2 \neq 0 \), which mathematically ensures that \( a \) and \( b \) are not both zero.

Sample Question 3. Any point on the y-axis is of the form
(a) (x, 0)
(b) (x, y)
(c) (0, y)
(d) (y, y)
Answer: (c) (0, y)
Every point located on the vertical y-axis has an x-coordinate (abscissa) equal to zero, while its y-coordinate (ordinate) can assume any real value.
In simple words: Along the y-axis, you do not move left or right, so the first coordinate is always zero.

Exam Tip: Keep this rule clear: points on the x-axis look like \( (x, 0) \), while points on the y-axis look like \( (0, y) \).

Exercise 4.1

Question 1. The linear equation 2x – 5y = 7 has
(a) A unique solution
(b) Two solutions
(c) Infinitely many solutions
(d) No solution
Answer: (c) Infinitely many solutions
Expressing \( x \) in terms of \( y \) gives \( x = \frac{7 + 5y}{2} \). Since \( y \) can take any real value from negative to positive infinity, there are infinitely many pairs \( (x, y) \) that satisfy this relationship.
In simple words: You can plug in any number you like for y and compute a matching x, giving infinite solutions.

Exam Tip: Unless a restricted domain such as integers or natural numbers is stated, every linear equation in two variables has infinitely many real solutions.

Question 2. The equation 2x + 5y = 7 has a unique solution, if x, y are :
(a) Natural numbers
(b) Positive real numbers
(c) Real numbers
(d) Rational numbers
Answer: (a) Natural numbers
Natural numbers must be positive integers: \( 1, 2, 3, \dots \).
If \( x = 1 \), then \( 2(1) + 5y = 7 \implies 5y = 5 \implies y = 1 \), which is a valid natural number.
If \( x = 2 \), \( 5y = 7 - 4 = 3 \implies y = \frac{3}{5} \) (not natural).
For any \( x \ge 3 \), \( 2x \ge 6 \), leaving \( 5y \le 1 \), so no other natural number solutions exist. Hence, \( (1, 1) \) is the sole solution.
In simple words: When restricted to positive counting numbers, only the pair x = 1 and y = 1 makes the equation work.

Exam Tip: Always pay close attention to the number set specified in the problem statement, as restrictions to natural numbers or integers drastically limit the solutions.

Question 3. If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is
(a) 4
(b) 6
(c) 5
(d) 2
Answer: (a) 4
Substitute the given coordinates \( x = 2 \) and \( y = 0 \) directly into the equation:
\( 2(2) + 3(0) = k \)
\( \implies 4 + 0 = k \)
\( \implies k = 4 \)
In simple words: Put 2 in place of x and 0 in place of y. Multiplying and adding gives k equal to 4.

Exam Tip: When testing if a coordinate point is a solution, replace the variables directly into the given equation and evaluate systematically.

Question 4. Any solution of the linear equation 2x + 0y + 9 = 0 in two variables is of the form
(a) (– 9/2 , m)
(b) (n, – 9/2)
(c) (0, – 9/2)
(d) (– 9, 0)
Answer: (a) (– 9/2 , m)
The given equation simplifies to \( 2x + 9 = 0 \), which implies \( x = -\frac{9}{2} \). Because the coefficient of \( y \) is zero, any real number choice for \( y \) (denoted here by \( m \)) leaves the equation completely true.
In simple words: The x-value is fixed at minus 9 over 2, while the y-value can be any real number m.

Exam Tip: A zero coefficient on \( y \) indicates that the line is vertical and parallel to the y-axis, meaning the x-coordinate stays constant for all points.

Question 5. The graph of the linear equation 2x + 3y = 6 cuts the y-axis at the point
(a) (2, 0)
(b) (0, 3)
(c) (3, 0)
(d) (0, 2)
Answer: (d) (0, 2)
Any line intersecting the vertical y-axis must have an abscissa of \( x = 0 \). Setting \( x = 0 \) gives:
\( 2(0) + 3y = 6 \implies 3y = 6 \implies y = 2 \)
Therefore, the required intersection coordinate is \( (0, 2) \).
In simple words: To find where a graph crosses the y-axis, set x to zero and solve for y.

Exam Tip: Never mix up intercepts: setting \( x = 0 \) finds the y-intercept, whereas setting \( y = 0 \) finds the x-intercept.

Question 6. The equation x = 7, in two variables, can be written as
(a) 1 . x + 1 . y = 7
(b) 1 . x + 0 . y = 7
(c) 0 . x + 1 . y = 7
(d) 0 . x + 0 . y = 7
Answer: (b) 1 . x + 0 . y = 7
To express a one-variable equation in the standard two-variable format \( ax + by = c \), introduce the missing variable with a coefficient of zero: \( 1 \cdot x + 0 \cdot y = 7 \).
In simple words: Since y does not appear, its coefficient must be zero while x keeps a coefficient of 1.

Exam Tip: Whenever converting a single-variable equation into two variables, always use zero as the coefficient of the absent variable.

Question 7. Any point on the x-axis is of the form
(a) (x, y)
(b) (0, y)
(c) (x, 0)
(d) (x, x)
Answer: (c) (x, 0)
Every point located on the horizontal x-axis has a vertical displacement of zero, which means its ordinate \( y \) is identically zero.
In simple words: On the x-axis, you stay flat on the horizontal line, meaning y is always zero.

Exam Tip: The equation of the x-axis itself is \( y = 0 \), so any point lying on it takes the form \( (x, 0) \).

Question 8. Any point on the line y = x is of the form
(a) (a, a)
(b) (0, a)
(c) (a, 0)
(d) (a, – a)
Answer: (a) (a, a)
The equation \( y = x \) mandates that the y-coordinate must match the x-coordinate exactly. Thus, choosing \( x = a \) leads directly to \( y = a \), yielding coordinates \( (a, a) \).
In simple words: Since y equals x, both numbers in the coordinate pair must be completely identical.

Exam Tip: The line \( y = x \) bisects the first and third quadrants at a \( 45^\circ \) angle to both coordinate axes.

Question 9. The equation of x-axis is of the form
(a) x = 0
(b) y = 0
(c) x + y = 0
(d) x = y
Answer: (b) y = 0
Because every point lying on the x-axis has an ordinate value of zero regardless of its abscissa, the algebraic equation representing the entire axis is \( y = 0 \).
In simple words: No matter where you stand on the x-axis, your height y is zero, so the equation is y = 0.

Exam Tip: Students frequently mix this up: the equation of the x-axis is \( y = 0 \), while the equation of the y-axis is \( x = 0 \).

Question 10. The graph of y = 6 is a line
(a) parallel to x-axis at a distance 6 units from the origin
(b) parallel to y-axis at a distance 6 units from the origin
(c) making an intercept 6 on the x-axis.
(d) making an intercept 6 on both the axes.
Answer: (a) parallel to x-axis at a distance 6 units from the origin
The equation \( y = 6 \) indicates that every point on the graph maintains a constant vertical distance of 6 units above the x-axis. Geometrically, this forms a horizontal straight line parallel to the x-axis.
In simple words: Since y stays fixed at 6 for all points, the line runs horizontally parallel to the x-axis at a distance of 6 units.

Exam Tip: Equations of the type \( y = k \) produce lines parallel to the x-axis, whereas equations of the type \( x = k \) produce lines parallel to the y-axis.

Question 11. x = 5, y = 2 is a solution of the linear equation
(a) x + 2 y = 7
(b) 5x + 2y = 7
(c) x + y = 7
(d) 5 x + y = 7
Answer: (c) x + y = 7
Testing the options with \( x = 5 \) and \( y = 2 \):
(a) \( 5 + 2(2) = 9 \ne 7 \)
(b) \( 5(5) + 2(2) = 29 \ne 7 \)
(c) \( 5 + 2 = 7 \), which matches the right-hand side.
(d) \( 5(5) + 2 = 27 \ne 7 \)
In simple words: Adding 5 and 2 gives 7, which makes option (c) the only true statement.

Exam Tip: Substitute the given values into each option one by one until the left-hand side matches the right-hand side.

Question 12. If a linear equation has solutions (–2, 2), (0, 0) and (2, – 2), then it is of the form
(a) y – x = 0
(b) x + y = 0
(c) –2x + y = 0
(d) –x + 2y = 0
Answer: (b) x + y = 0
Observe the sum of coordinates for each given pair:
\( -2 + 2 = 0 \), \( 0 + 0 = 0 \), and \( 2 + (-2) = 0 \).
Each point satisfies the relation \( x + y = 0 \).
In simple words: In every given pair, adding the x and y values together equals zero, so the equation is x + y = 0.

Exam Tip: Look for immediate arithmetic patterns between coordinates, such as \( x + y = 0 \) or \( y = -x \), to quickly identify the matching equation.

Question 13. The positive solutions of the equation ax + by + c = 0 always lie in the
(a) 1st quadrant
(b) 2nd quadrant
(c) 3rd quadrant
(d) 4th quadrant
Answer: (a) 1st quadrant
A positive solution requires that both variables are strictly greater than zero, meaning \( x > 0 \) and \( y > 0 \). Points with both coordinates positive reside exclusively in the first quadrant.
In simple words: Positive values for both x and y always belong to the top-right, which is the 1st quadrant.

Exam Tip: Remember the quadrant sign conventions: 1st (+, +), 2nd (-, +), 3rd (-, -), and 4th (+, -).

Question 14. The graph of the linear equation 2x + 3y = 6 is a line which meets the x-axis at the point
(a) (0, 2)
(b) (2, 0)
(c) (3, 0)
(d) (0, 3)
Answer: (c) (3, 0)
Points on the x-axis have an ordinate of \( y = 0 \). Setting \( y = 0 \) in the equation gives:
\( 2x + 3(0) = 6 \implies 2x = 6 \implies x = 3 \)
Thus, the line cuts the x-axis at the coordinate \( (3, 0) \).
In simple words: Substitute zero for y and divide 6 by 2 to get x = 3, giving the point (3, 0).

Exam Tip: Points intersecting the x-axis must always end with 0 as their second number: \( (x, 0) \).

Question 15. The graph of the linear equation y = x passes through the point
(a) \( \left(\frac{3}{2}, -\frac{3}{2}\right) \)
(b) \( \left(0, \frac{3}{2}\right) \)
(c) (1, 1)
(d) \( \left(-\frac{1}{2}, \frac{1}{2}\right) \)
Answer: (c) (1, 1)
The equation \( y = x \) requires that the abscissa and the ordinate must be numerically identical and share the same sign. Among the choices, only \( (1, 1) \) satisfies \( y = x \).
In simple words: The line y = x only includes points where both coordinates are exactly the same number.

Exam Tip: Do not choose coordinates with opposite signs like \( \left(\frac{3}{2}, -\frac{3}{2}\right) \); those belong to the line \( y = -x \).

Question 16. If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation :
(a) Changes
(b) Remains the same
(c) Changes in case of multiplication only
(d) Changes in case of division only
Answer: (b) Remains the same
Multiplying or dividing both members of an equation by any non-zero real number yields an equivalent equation whose solution set is completely identical to the original.
In simple words: Scaling both sides of an equation by a non-zero number keeps the balance intact, so the answers do not change.

Exam Tip: Non-zero multiplication or division creates an equivalent linear equation with the exact same geometric line as its graph.

Question 17. How many linear equations in x and y can be satisfied by x = 1 and y = 2?
(a) Only one
(b) Two
(c) Infinitely many
(d) Three
Answer: (c) Infinitely many
A solution \( x = 1, y = 2 \) corresponds to the single geometric point \( (1, 2) \) in the plane. Because infinitely many straight lines can pass through any single given point, there exist infinitely many linear equations satisfied by these coordinates.
In simple words: Countless different lines can pass through the point (1, 2), so infinitely many equations can have it as a solution.

Exam Tip: Euclid's axiom states that an infinite number of lines pass through a single point; each such line represents a unique linear equation.

Question 18. The point of the form (a, a) always lies on :
(a) x-axis
(b) y-axis
(c) On the line y = x
(d) On the line x + y = 0
Answer: (c) On the line y = x
Because the point has an x-coordinate of \( a \) and a y-coordinate of \( a \), the condition \( y = x \) is fulfilled for every real value of \( a \).
In simple words: When both coordinates are equal, the point must sit on the line y = x.

Exam Tip: Identify the algebraic relationship between coordinates: identical values indicate the line \( y = x \).

Question 19. The point of the form (a, – a) always lies on the line
(a) x = a
(b) y = – a
(c) y = x
(d) x + y = 0
Answer: (d) x + y = 0
Adding the coordinates together gives \( x + y = a + (-a) = 0 \). Hence, any point of the form \( (a, -a) \) always satisfies the equation \( x + y = 0 \).
In simple words: A number plus its opposite equals zero, so any pair like (a, -a) lies on x + y = 0.

Exam Tip: Notice that \( y = -x \) is algebraically identical to \( x + y = 0 \).

(C) Short Answer Questions with Reasoning

Sample Question 1. Write whether the following statements are True or False? Justify your answers.
(i) ax + by + c = 0, where a, b and c are real numbers, is a linear equation in two variables.
(ii) A linear equation 2x + 3y = 5 has a unique solution.
(iii) All the points (2, 0), (–3, 0), (4, 2) and (0, 5) lie on the x-axis.
(iv) The line parallel to the y-axis at a distance 4 units to the left of y-axis is given by the equation x = – 4.
(v) The graph of the equation y = mx + c passes through the origin.
Answer:
(i) False. The equation represents a linear equation in two variables only if both coefficients \( a \) and \( b \) are non-zero. If either is zero, it simplifies to an equation in a single variable.
(ii) False. A linear equation in two variables graphed over real numbers has infinitely many solutions rather than a unique one.
(iii) False. Only points with an ordinate of zero lie on the x-axis, which holds for \( (2, 0) \) and \( (-3, 0) \). Meanwhile, \( (4, 2) \) lies in the first quadrant and \( (0, 5) \) lies on the y-axis.
(iv) True. Any vertical line positioned \( a \) units to the left of the y-axis is represented by the equation \( x = -a \). For a distance of 4 units, the equation is \( x = -4 \).
(v) False. Testing the origin \( (0, 0) \) yields \( 0 = m(0) + c \implies c = 0 \). The graph only passes through the origin when the constant term \( c \) is equal to zero.
In simple words: Make sure variables have non-zero multipliers, remember single equations have infinite solutions, check if coordinates have zeros in the right spot, and test (0, 0) to see if lines pass through the origin.

Exam Tip: For True/False questions with reasoning, always provide a specific mathematical counterexample whenever stating "False".

Sample Question 2. Write whether the following statement is True or False? Justify your answer.
The coordinates of points given in the table :

represent some of the solutions of the equation 2x + 2 = y.
Answer: True. Evaluating the linear equation \( y = 2x + 2 \) for each table entry shows that every point matches:
- For \( x = 0 \): \( y = 2(0) + 2 = 2 \)
- For \( x = 1 \): \( y = 2(1) + 2 = 4 \)
- For \( x = 2 \): \( y = 2(2) + 2 = 6 \)
- For \( x = 3 \): \( y = 2(3) + 2 = 8 \)
- For \( x = 4 \): \( y = 2(4) + 2 = 10 \)
Each computed y-value is precisely two units greater than twice the corresponding x-value.
In simple words: Doubling each x and adding 2 gives the exact y shown in the table for every single pair.

Exam Tip: Verify every column in the table individually; a statement is false if even one pair fails to satisfy the equation.

Exercise 4.2

Question 1. The point (0, 3) lies on the graph of the linear equation 3x + 4y = 12.
Answer: True.
Substitute \( x = 0 \) and \( y = 3 \) into the left-hand side:
\( \text{L.H.S.} = 3(0) + 4(3) = 0 + 12 = 12 = \text{R.H.S.} \)
Because the point satisfies the given equation, it lies on the line.
In simple words: Putting 0 and 3 into the equation gives 12, which matches the right side completely.

Exam Tip: Showing that LHS equals RHS is mandatory to receive full justification marks.

Question 2. The graph of the linear equation x + 2y = 7 passes through the point (0, 7).
Answer: False.
Substitute \( x = 0 \) and \( y = 7 \) into the left-hand side:
\( \text{L.H.S.} = 0 + 2(7) = 14 \ne 7 \)
Because the left side does not equal the right side, the point \( (0, 7) \) does not lie on the graph.
In simple words: When you plug in 0 and 7, you get 14, not 7, so the point is not on the line.

Exam Tip: Clearly write \( \text{L.H.S.} \ne \text{R.H.S.} \) with the calculated numeric value when proving a statement is False.

Question 3. The graph given below represents the linear equation x + y = 0.
Answer: True.
The points highlighted on the graph are \( (-3, 3) \), \( (-1, 1) \), and the origin \( (0, 0) \).
Testing these coordinates in the equation \( x + y = 0 \):
- For \( (-3, 3) \): \( -3 + 3 = 0 \)
- For \( (-1, 1) \): \( -1 + 1 = 0 \)
- For \( (0, 0) \): \( 0 + 0 = 0 \)
Because every point shown on the plotted line satisfies \( x + y = 0 \), the graph accurately depicts this equation.
In simple words: All points on the line add up to zero, confirming that the line shows x + y = 0.

Exam Tip: Select at least two distinct points from the given figure and demonstrate that both satisfy the linear equation.

Question 4. The graph given below represents the linear equation x = 3 (see Fig. 4.2).
Answer: True.
The graph illustrates a vertical straight line parallel to the y-axis, located at a constant perpendicular distance of 3 units to the right of the origin. Every point on this line has an x-coordinate of 3, which is the exact geometric definition of \( x = 3 \).
In simple words: The line runs vertically straight through 3 on the x-axis, which is exactly what x = 3 looks like.

Exam Tip: A straight line parallel to the y-axis at distance \( c \) is always represented by the equation \( x = c \).

Question 5. The coordinates of points in the table:

represent some of the solutions of the equation x – y + 2 = 0.
Answer: False.
Check whether the point \( (3, -5) \) satisfies the equation \( x - y + 2 = 0 \):
\( \text{L.H.S.} = 3 - (-5) + 2 = 3 + 5 + 2 = 10 \ne 0 \)
Since \( (3, -5) \) does not satisfy the equation, the table does not represent solutions of \( x - y + 2 = 0 \).
In simple words: The fourth point gives 10 instead of 0, so the table fails.

Exam Tip: Even if four out of five points satisfy the equation, a single failing point makes the entire statement False.

Question 6. Every point on the graph of a linear equation in two variables does not represent a solution of the linear equation.
Answer: False.
By fundamental definition, every single point situated on the graph of a linear equation represents a valid solution of that equation, and conversely, every solution corresponds to a point on the line.
In simple words: Every point on the line is a solution, with no exceptions.

Exam Tip: The correspondence between the points on a line and the solutions of its linear equation is strictly one-to-one.

Question 7. The graph of every linear equation in two variables need not be a line.
Answer: False.
The geometric representation of every first-degree (linear) polynomial equation in two variables is invariably a straight line in the Cartesian coordinate system.
In simple words: The graph of a linear equation is always a straight line without exception.

Exam Tip: The word "linear" originates directly from the geometric fact that its graph forms a straight line.

(D) Short Answer Questions

Sample Question 1. Find the points where the graph of the equation 3x + 4y = 12 cuts the x-axis and the y-axis.
Answer: To determine the x-intercept, substitute \( y = 0 \) into the equation:
\( 3x + 4(0) = 12 \implies 3x = 12 \implies x = 4 \)
So, the graph crosses the x-axis at the coordinate \( (4, 0) \).
To determine the y-intercept, substitute \( x = 0 \) into the equation:
\( 3(0) + 4y = 12 \implies 4y = 12 \implies y = 3 \)
So, the graph crosses the y-axis at the coordinate \( (0, 3) \).
In simple words: Set y = 0 to get x = 4, and set x = 0 to get y = 3. The line cuts the axes at (4, 0) and (0, 3).

Exam Tip: State both points clearly as ordered pairs \( (x, y) \); writing only \( x = 4 \) and \( y = 3 \) may result in a loss of presentation marks.

Sample Question 2. At what point does the graph of the linear equation x + y = 5 meet a line which is parallel to the y-axis, at a distance 2 units from the origin and in the positive direction of x-axis.
Answer: A line running parallel to the y-axis at a positive distance of 2 units from the origin has the fixed equation \( x = 2 \).
Substitute \( x = 2 \) into the given linear equation:
\( 2 + y = 5 \implies y = 5 - 2 = 3 \)
Hence, the two lines intersect at the point \( (2, 3) \).
In simple words: The vertical line has x = 2 everywhere. Plugging 2 into x + y = 5 gives y = 3, so they meet at (2, 3).

Exam Tip: First translate the verbal geometric description into the equation \( x = 2 \) before substituting into the given equation.

Sample Question 3. Determine the point on the graph of the equation 2x + 5y = 20 whose x-coordinate is 5/2 times its ordinate.
Answer: Let the ordinate (y-coordinate) be \( y \). The problem specifies that the abscissa is:
\( x = \frac{5}{2}y \)
Substitute this expression for \( x \) into the given equation:
\( 2\left(\frac{5}{2}y\right) + 5y = 20 \)
\( \implies 5y + 5y = 20 \)
\( \implies 10y = 20 \implies y = 2 \)
Now determine the corresponding value of \( x \):
\( x = \frac{5}{2}(2) = 5 \)
Therefore, the required coordinate on the graph is \( (5, 2) \).
In simple words: Replace x with 5/2 y. That simplifies the equation to 10y = 20, giving y = 2 and x = 5.

Exam Tip: Remember: abscissa means the x-coordinate, and ordinate means the y-coordinate; do not reverse them.

Sample Question 4. Draw the graph of the equation represented by the straight line which is parallel to the x-axis and is 4 units above it.
Answer: Any line situated parallel to the horizontal x-axis at an upward distance of \( k \) units has the form \( y = k \). With \( k = 4 \), the equation is \( y = 4 \).
To draw the line, identify any two points having an ordinate of 4, such as \( (1, 4) \) and \( (2, 4) \). Plotting these points and connecting them produces the horizontal line \( y = 4 \).
In simple words: The equation is y = 4. Draw a flat horizontal line 4 units above the x-axis passing through (1, 4) and (2, 4).

Exam Tip: Clearly label at least two plotted points on the graph line along with the equation of the line.

Exercise 4.3

Question 1. Draw the graphs of linear equations y = x and y = – x on the same cartesian plane. What do you observe?
Answer: For the line \( y = x \), choose coordinate pairs: \( (0, 0) \), \( (2, 2) \), and \( (-2, -2) \).
For the line \( y = -x \), choose coordinate pairs: \( (0, 0) \), \( (2, -2) \), and \( (-2, 2) \).
Plotting both sets of points on the same Cartesian coordinate system produces two intersecting lines.
Observation: Both lines pass directly through the origin \( (0, 0) \) and cross each other at right angles (\( 90^\circ \)).
In simple words: Both graphs intersect at the origin (0, 0) and are perpendicular to each other.

Exam Tip: Be sure to write down the explicit geometric observation: the lines are perpendicular and cross at the origin.

Question 2. Determine the point on the graph of the linear equation 2x + 5y = 19, whose ordinate is 1 1/2 times its abscissa.
Answer: Let the abscissa be \( x \). The ordinate is given by:
\( y = 1\frac{1}{2}x = \frac{3}{2}x \)
Substitute \( y = \frac{3}{2}x \) into the equation:
\( 2x + 5\left(\frac{3}{2}x\right) = 19 \)
\( \implies 2x + \frac{15}{2}x = 19 \)
\( \implies \frac{19}{2}x = 19 \)
\( \implies x = 2 \)
Now compute the ordinate \( y \):
\( y = \frac{3}{2}(2) = 3 \)
Therefore, the required coordinate point is \( (2, 3) \).
In simple words: Set y equal to 3/2 x. Solving 19/2 x = 19 gives x = 2, which then makes y = 3.

Exam Tip: Convert mixed numbers like \( 1\frac{1}{2} \) into improper fractions (\( \frac{3}{2} \)) right away to simplify algebraic manipulation.

Question 3. Draw the graph of the equation represented by a straight line which is parallel to the x-axis and at a distance 3 units below it.
Answer: A horizontal line situated at a distance of 3 units below the x-axis maintains a constant vertical coordinate of \( y = -3 \).
The linear equation is therefore \( y = -3 \).
To construct the graph, plot points with ordinate \( -3 \), such as \( (-2, -3) \), \( (0, -3) \), and \( (2, -3) \), and join them with a straight line.
In simple words: The equation is y = -3. Draw a horizontal straight line 3 units below the x-axis.

Exam Tip: "Below the x-axis" means the y-coordinate is negative, so the constant is \( -3 \), not \( +3 \).

Question 4. Draw the graph of the linear equation whose solutions are represented by the points having the sum of the coordinates as 10 units.
Answer: Let the coordinates of any solution point be \( (x, y) \).
The condition states that the sum of the coordinates is 10:
\( x + y = 10 \)
To graph this equation, find at least two coordinate pairs:
- For \( x = 0 \): \( y = 10 \implies (0, 10) \)
- For \( y = 0 \): \( x = 10 \implies (10, 0) \)
- For \( x = 5 \): \( y = 5 \implies (5, 5) \)
Plotting these points and joining them with a straight line forms the graph.
In simple words: The equation is x + y = 10. Connect (0, 10) on the y-axis to (10, 0) on the x-axis to draw the line.

Exam Tip: First translate the word problem into the algebraic equation \( x + y = 10 \) before making the table of values.

Question 5. Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa.
Answer: Let the abscissa (x-coordinate) of a general point be \( x \), and its ordinate (y-coordinate) be \( y \).
The given condition requires that the ordinate is 3 times the abscissa:
\( y = 3x \)
In standard linear equation form, this is written as:
\( 3x - y = 0 \)
In simple words: The y-coordinate is three times the x-coordinate, which gives the equation y = 3x.

Exam Tip: State the equation in standard form \( ax + by + c = 0 \), i.e., \( 3x - y = 0 \), alongside \( y = 3x \).

Question 6. If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a.
Answer: Since the point \( (3, 4) \) lies on the graph, its coordinates must satisfy the equation.
Substitute \( x = 3 \) and \( y = 4 \) into \( 3y = ax + 7 \):
\( 3(4) = a(3) + 7 \)
\( \implies 12 = 3a + 7 \)
\( \implies 3a = 12 - 7 \)
\( \implies 3a = 5 \)
\( \implies a = \frac{5}{3} \)
In simple words: Replace x with 3 and y with 4. Subtract 7 from 12 to get 5, then divide by 3 to find a = 5/3.

Exam Tip: Leaving the answer as an improper fraction \( \frac{5}{3} \) is completely acceptable; do not round to decimals unless instructed.

Question 7. How many solution(s) of the equation 2x + 1 = x – 3 are there on the :
(i) Number line
(ii) Cartesian plane
Answer: First, simplify the given linear equation:
\( 2x - x = -3 - 1 \implies x = -4 \)
(i) On the number line, \( x = -4 \) represents a single point. Hence, there is a unique (only one) solution.
(ii) On the Cartesian plane, \( x = -4 \) is a linear equation in two variables (\( 1\cdot x + 0\cdot y = -4 \)), which describes a vertical line parallel to the y-axis. Because every point lying on this vertical line is a solution, there are infinitely many solutions.
In simple words: On a 1D number line, x = -4 is just one dot. On a 2D grid, it becomes an entire vertical line containing infinite points.

Exam Tip: Make sure to distinguish between the 1-dimensional representation (a unique point) and the 2-dimensional representation (a line with infinitely many points).

Question 8. Find the solution of the linear equation x + 2y = 8 which represents a point on
(i) x-axis
(ii) y-axis
Answer:
(i) For a point on the x-axis, set \( y = 0 \):
\( x + 2(0) = 8 \implies x = 8 \)
Thus, the point is \( (8, 0) \).
(ii) For a point on the y-axis, set \( x = 0 \):
\( 0 + 2y = 8 \implies 2y = 8 \implies y = 4 \)
Thus, the point is \( (0, 4) \).
In simple words: Setting y = 0 gives (8, 0) on the x-axis. Setting x = 0 gives (0, 4) on the y-axis.

Exam Tip: Present each answer as an ordered pair \( (x, y) \) rather than writing only single numeric values.

Question 9. For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution.
Answer: Because the values of \( x \) and \( y \) are equal in the solution, we have \( y = x \).
Substitute \( y = x \) into the given equation:
\( 2x + c(x) = 8 \)
\( \implies cx = 8 - 2x \)
Dividing by \( x \) (for any non-zero value of \( x \)):
\( \implies c = \frac{8 - 2x}{x}, \quad x \ne 0 \)
(Alternatively, any real number \( c \ne -2 \) allows equal values of \( x \) and \( y \), given by \( x = y = \frac{8}{2 + c} \)).
In simple words: Replace y with x to get (2 + c)x = 8, which means c equals (8 - 2x)/x for any non-zero x.

Exam Tip: Mention the condition \( x \neq 0 \) when expressing \( c \) in terms of \( x \), or note that \( c \neq -2 \) ensures a finite solution.

Question 10. Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is the value of y when x = 5?
Answer: Because \( y \) varies directly as \( x \), their relationship is modeled by \( y = kx \), where \( k \) is a non-zero constant of proportionality.
Substitute \( x = 4 \) and \( y = 12 \) to find \( k \):
\( 12 = k(4) \implies k = \frac{12}{4} = 3 \)
Therefore, the linear equation is:
\( y = 3x \)
Now, when \( x = 5 \):
\( y = 3(5) = 15 \)
In simple words: 12 divided by 4 gives a constant ratio of 3, so the formula is y = 3x. For x = 5, y is 3 times 5, which equals 15.

Exam Tip: Remember to write down both answers asked in the problem: the linear equation (\( y = 3x \)) and the calculated value (\( y = 15 \)).

(E) Long Answer Questions

Sample Question 1. Draw the graph of the linear equation 2x + 3y = 12. At what points, the graph of the equation cuts the x-axis and the y-axis?
Answer: The given equation is \( 2x + 3y = 12 \).
Expressing \( y \) in terms of \( x \):
\( y = \frac{12 - 2x}{3} \)
Find two convenient points to draw the line:
- For \( x = 0 \): \( y = \frac{12}{3} = 4 \implies A(0, 4) \) lies on the graph.
- For \( y = 0 \): \( 2x = 12 \implies x = 6 \implies B(6, 0) \) lies on the graph.
Plot points \( A(0, 4) \) and \( B(6, 0) \) and connect them with a straight line.
From the graph, the line intersects the x-axis at \( (6, 0) \) and the y-axis at \( (0, 4) \).
In simple words: The graph crosses the x-axis at (6, 0) and the y-axis at (0, 4).

Exam Tip: Finding the two axis intercepts is the fastest and most accurate method to plot any linear equation in two variables.

Sample Question 2. The following values of x and y are thought to satisfy a linear equation :

Draw the graph, using the values of x, y as given in the above table.
At what point the graph of the linear equation
(i) cuts the x-axis.
(ii) cuts the y-axis.
Answer: Plot the two coordinate points \( A(1, 1) \) and \( B(2, 3) \) and draw the extended straight line passing through both.
The slope of the line is \( m = \frac{3 - 1}{2 - 1} = 2 \), giving the equation \( y - 1 = 2(x - 1) \implies y = 2x - 1 \).
(i) The graph cuts the x-axis at \( \left(\frac{1}{2}, 0\right) \).
(ii) The graph cuts the y-axis at \( (0, -1) \).
In simple words: The line passes through (1, 1) and (2, 3), crossing the x-axis at (1/2, 0) and the y-axis at (0, -1).

Exam Tip: Extend the line with a ruler across both axes to visually read and verify the intercept coordinates accurately.

Sample Question 3. The Autorikshaw fare in a city is charged Rs 10 for the first kilometer and @ Rs 4 per kilometer for subsequent distance covered. Write the linear equation to express the above statement. Draw the graph of the linear equation.
Answer: Let the total distance traveled be \( x\text{ km} \) and the total fare charged be Rs \( y \).
The fare for the first kilometer is Rs 10.
The remaining distance is \( (x - 1)\text{ km} \), charged at the rate of Rs 4 per kilometer:
\( y = 10 + 4(x - 1) \)
\( \implies y = 10 + 4x - 4 \)
\( \implies y = 4x + 6 \)
To draw the graph, select points:
- For \( x = 0 \): \( y = 6 \implies (0, 6) \)
- For \( x = -1 \): \( y = 2 \implies (-1, 2) \)
- For \( x = 1 \): \( y = 10 \implies (1, 10) \)
In simple words: The fare is 10 for the first km plus 4 for every additional km, giving the linear equation y = 4x + 6.

Exam Tip: Be sure to subtract 1 from \( x \) for the subsequent distance: \( 4(x - 1) \); writing \( 10 + 4x \) is a very common error.

Sample Question 4. The work done by a body on application of a constant force is the product of the constant force and the distance travelled by the body in the direction of force. Express this in the form of a linear equation in two variables and draw its graph by taking the constant force as 3 units. What is the work done when the distance travelled is 2 units. Verify it by plotting the graph.
Answer: Let the distance covered be \( x \) units and the work done be \( y \) units.
Given that \( \text{Work done} = (\text{constant force}) \times (\text{distance}) \), and the force is 3 units:
\( y = 3x \)
When the distance traveled is \( x = 2 \) units:
\( y = 3(2) = 6\text{ units} \)
To draw the graph of \( y = 3x \), plot points \( A(0, 0) \), \( B(1, 3) \), and \( C(2, 6) \). Since distance and work cannot be negative in this physical scenario, the ray originates from \( (0, 0) \) into the first quadrant.
Verification from Graph: Drawing a perpendicular line from \( x = 2 \) on the x-axis meets the graph at point \( C(2, 6) \). Projecting horizontally from \( C \) to the y-axis confirms that \( y = 6 \) units.
In simple words: The formula is y = 3x. For 2 units of distance, the graph reaches a height of 6 units, verifying that work done is 6 units.

Exam Tip: Draw dashed projection lines from \( x = 2 \) up to the graph and across to \( y = 6 \) on the y-axis to show clear graphical verification.

Exercise 4.4

Question 1. 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.
Answer: Substitute the coordinates of each point into the equation \( y = 9x - 7 \):
- For point \( A(1, 2) \):
\( \text{R.H.S.} = 9(1) - 7 = 9 - 7 = 2 = y = \text{L.H.S.} \)
- For point \( B(-1, -16) \):
\( \text{R.H.S.} = 9(-1) - 7 = -9 - 7 = -16 = y = \text{L.H.S.} \)
- For point \( C(0, -7) \):
\( \text{R.H.S.} = 9(0) - 7 = 0 - 7 = -7 = y = \text{L.H.S.} \)
Because every point satisfies the linear equation, points \( A \), \( B \), and \( C \) all lie on its graph.
In simple words: Plugging in the x-values gives the exact matching y-values for all three points, proving they all sit on the line.

Exam Tip: Verify each given point separately with a dedicated step to ensure complete marks for proof questions.

Question 2. The following observed values of x and y are thought to satisfy a linear equation. Write the linear equation :

Draw the graph using the values of x, y as given in the above table.
At what points the graph of the linear equation
(i) cuts the x-axis
(ii) cuts the y-axis
Answer: Let the linear equation be \( ax + by + c = 0 \). Using the two points \( (6, -2) \) and \( (-6, 6) \):
Slope \( m = \frac{6 - (-2)}{-6 - 6} = \frac{8}{-12} = -\frac{2}{3} \)
Using point-slope form:
\( y - (-2) = -\frac{2}{3}(x - 6) \)
\( \implies 3(y + 2) = -2(x - 6) \)
\( \implies 3y + 6 = -2x + 12 \)
\( \implies 2x + 3y = 6 \)
Plotting the points \( (6, -2) \) and \( (-6, 6) \) gives the graph of the straight line.
(i) Setting \( y = 0 \) in \( 2x + 3(0) = 6 \implies x = 3 \). The graph cuts the x-axis at \( (3, 0) \).
(ii) Setting \( x = 0 \) in \( 2(0) + 3y = 6 \implies y = 2 \). The graph cuts the y-axis at \( (0, 2) \).
In simple words: The equation connecting the points is 2x + 3y = 6. It crosses the horizontal axis at (3, 0) and the vertical axis at (0, 2).

Exam Tip: First use the two given table points to derive the linear equation \( 2x + 3y = 6 \) before calculating the intercepts.

Question 3. Draw the graph of the linear equation 3x + 4y = 6. At what points, the graph cuts the x-axis and the y-axis.
Answer: For the given equation \( 3x + 4y = 6 \):
- To find the x-intercept, set \( y = 0 \):
\( 3x + 4(0) = 6 \implies 3x = 6 \implies x = 2 \)
The graph intersects the x-axis at \( (2, 0) \).
- To find the y-intercept, set \( x = 0 \):
\( 3(0) + 4y = 6 \implies 4y = 6 \implies y = \frac{6}{4} = \frac{3}{2} = 1.5 \)
The graph intersects the y-axis at \( (0, 1.5) \) or \( \left(0, \frac{3}{2}\right) \).
Plot points \( (2, 0) \) and \( (0, 1.5) \) to draw the line.
In simple words: Setting y = 0 gives (2, 0) on the x-axis, and setting x = 0 gives (0, 1.5) on the y-axis.

Exam Tip: Fractional intercepts like \( 1.5 \) can be accurately marked on graph paper by choosing an appropriate scale such as 2 divisions = 1 unit.

Question 4. The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation
C = (5F - 160) / 9
(i) If the temperature is 86°F, what is the temperature in Celsius?
(ii) If the temperature is 35°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 0°C what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
(iv) What is the numerical value of the temperature which is same in both the scales?
Answer:
(i) Substitute \( F = 86 \):
\( C = \frac{5(86) - 160}{9} = \frac{430 - 160}{9} = \frac{270}{9} = 30^\circ\text{C} \)

(ii) Substitute \( C = 35 \):
\( 35 = \frac{5F - 160}{9} \implies 315 = 5F - 160 \implies 5F = 475 \implies F = 95^\circ\text{F} \)

(iii) For \( C = 0^\circ\text{C} \):
\( 0 = \frac{5F - 160}{9} \implies 5F - 160 = 0 \implies 5F = 160 \implies F = 32^\circ\text{F} \)
For \( F = 0^\circ\text{F} \):
\( C = \frac{5(0) - 160}{9} = -\frac{160}{9}^\circ\text{C} \approx -17.8^\circ\text{C} \)

(iv) Let the common numerical reading be \( x \), so \( C = F = x \):
\( x = \frac{5x - 160}{9} \)
\( \implies 9x = 5x - 160 \)
\( \implies 4x = -160 \)
\( \implies x = -40 \)
The numerical temperature that is identical in both temperature scales is \( -40^\circ \).
In simple words: 86°F is 30°C, 35°C is 95°F, 0°C is 32°F, 0°F is -160/9°C, and both thermometers show the exact same reading at -40°.

Exam Tip: For part (iv), setting \( C = F = x \) and solving the simple linear equation directly scores full credit without trial and error.

Question 5. If the temperature of a liquid can be measured in kelvin units as x°K or in Fahrenheit units as y°F. The relation between the two systems of measurement of temperature is given by the linear equation
y = 9/5 (x – 273) + 32
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the body is 313°K.
(ii) If the temperature is 158° F, then find the temperature in Kelvin.
Answer:
(i) Substitute \( x = 313 \) into the formula:
\( y = \frac{9}{5}(313 - 273) + 32 \)
\( \implies y = \frac{9}{5}(40) + 32 \)
\( \implies y = 9(8) + 32 = 72 + 32 = 104^\circ\text{F} \)

(ii) Substitute \( y = 158 \) into the formula:
\( 158 = \frac{9}{5}(x - 273) + 32 \)
\( \implies 158 - 32 = \frac{9}{5}(x - 273) \)
\( \implies 126 = \frac{9}{5}(x - 273) \)
Multiply by \( \frac{5}{9} \):
\( \implies x - 273 = \frac{126 \times 5}{9} = 14 \times 5 = 70 \)
\( \implies x = 70 + 273 = 343\text{ K} \)
In simple words: 313 K converts to 104°F, and 158°F converts back to 343 K.

Exam Tip: Subtract 32 first before multiplying by \( \frac{5}{9} \) when converting from Fahrenheit back to Kelvin.

Question 6. The force exerted to pull a cart is directly proportional to the acceleration produced in the body. Express the statement as a linear equation of two variables and draw the graph of the same by taking the constant mass equal to 6 kg. Read from the graph, the force required when the acceleration produced is (i) 5 m/sec2 , (ii) 6 m/sec2 .
Answer: Let the applied force be denoted by \( y \) (in Newtons) and the resulting acceleration by \( x \) (in \( \text{m/s}^2 \)).
By Newton's second law, \( \text{Force} = \text{mass} \times \text{acceleration} \). With constant mass \( m = 6\text{ kg} \):
\( y = 6x \)
Points to plot the linear equation:
- For \( x = 0 \): \( y = 0 \implies (0, 0) \)
- For \( x = 1 \): \( y = 6 \implies (1, 6) \)
- For \( x = 2 \): \( y = 12 \implies (2, 12) \)
Reading directly from the graph:
(i) When acceleration \( x = 5\text{ m/sec}^2 \), the vertical height gives force \( y = 30\text{ N} \).
(ii) When acceleration \( x = 6\text{ m/sec}^2 \), the vertical height gives force \( y = 36\text{ N} \).
In simple words: The equation is y = 6x. Looking up 5 on the horizontal axis gives 30 N of force, and looking up 6 gives 36 N of force.

Exam Tip: Always include the appropriate physical units (Newtons / N and m/sec²) alongside your numerical values.