CBSE Class 9 Mathematics Revision Worksheet Set 04

Q1. What is the form of the point on the y-axis?
Answer: The x-coordinate of any point on the y-axis is always zero.
Therefore, the form of any point on the y-axis is \((0, y)\), where y is a real number.
In simple words: Any point on the vertical y-axis has a zero horizontal distance, so its coordinate is always of the form \((0, y)\).

Exam Tip: Remember that for points on the x-axis, y is zero \((x, 0)\), and for points on the y-axis, x is zero \((0, y)\).

Q2. If \((2, 0)\) is a solution of the linear equation \(2x + 3y = k\), then what is the value of k?
Answer: Given equation is \(2x + 3y = k\)
Since \((2, 0)\) is a solution, substitute \(x = 2\) and \(y = 0\) into the equation:
\(2(2) + 3(0) = k\)
\(4 + 0 = k\)
\(k = 4\)
In simple words: Substitute the given x and y values into the equation to find the unknown constant k.

Exam Tip: A solution to an equation always satisfies the equation when the coordinates are substituted for x and y.

Q3. What is the form of any solution of the linear equation \(2x + 0y + 9 = 0\)?
Answer: Given equation: \(2x + 9 = 0\)
Solving for x:
\(2x = -9\)
\(x = -\frac{9}{2} = -4.5\)
Since there is no restriction on y, y can be any real number.
Therefore, any solution is of the form \(\left(-\frac{9}{2}, y\right)\) or \((-4.5, y)\), where y is any real number.
In simple words: The x-coordinate is fixed at -4.5, but the y-coordinate can be anything.

Exam Tip: When the coefficient of a variable is zero, that variable can take any arbitrary real value.

Q4. At what point does the graph \(2x + 3y = 6\) cuts the y-axis?
Answer: To find where the graph cuts the y-axis, put \(x = 0\) in the equation \(2x + 3y = 6\):
\(2(0) + 3y = 6\)
\(3y = 6\)
\(y = \frac{6}{3} = 2\)
Thus, the graph cuts the y-axis at the point \((0, 2)\).
In simple words: Set x to zero to find the y-intercept.

Exam Tip: To find the y-intercept of any line, always substitute \(x = 0\).

Q5. How can you write \(x = 7\) in two variables?
Answer: The equation \(x = 7\) in one variable can be expressed in two variables x and y as:
\(1x + 0y - 7 = 0\) or \(1x + 0y = 7\).
In simple words: Add a zero coefficient term for y to write a single-variable equation in two variables.

Q6. What will be the form of any point on the line \(y = x\)?
Answer: On the line \(y = x\), the y-coordinate is always equal to the x-coordinate.
Therefore, any point on this line is of the form \((a, a)\), where a is a real number.
In simple words: Both coordinates are identical for every point on the line \(y = x\).

Exam Tip: This line passes through the origin and makes a 45-degree angle with the positive axes.

Q7. What is the equation of (i) x-axis (ii) y-axis?
Answer:
(i) The equation of the x-axis is \(y = 0\).
(ii) The equation of the y-axis is \(x = 0\).
In simple words: On the x-axis y is always 0, and on the y-axis x is always 0.

Exam Tip: Memorize these fundamental axis equations as they are frequently used in coordinate geometry.

Q8. The graph of \(y = 6\) is a line parallel to ________?
Answer: The graph of \(y = 6\) is a line parallel to the x-axis at a distance of 6 units above it.
In simple words: A constant y value forms a horizontal line parallel to the x-axis.

Exam Tip: Lines of the form \(y = k\) are horizontal and parallel to the x-axis, while \(x = k\) are vertical and parallel to the y-axis.

Q9. If we multiply or divide both sides of a linear equation with a non-zero number, then what will happen to the solution?
Answer: The solution of the linear equation remains unchanged (the roots or solutions are unaffected).
In simple words: Multiplying or dividing an equation by a non-zero number gives an equivalent equation with the exact same solution.

Q10. How many linear equations in x and y can be satisfied by \(x = 1\) and \(y = 2\)?
Answer: Infinitely many linear equations can be satisfied by a single pair of values \((1, 2)\).
In simple words: An infinite number of straight lines pass through a single point \((1, 2)\).

Exam Tip: While a single point allows infinitely many linear equations, a unique linear equation requires at least two distinct points.

Q11. Where will the point of the form \((a, a)\) lie?
Answer: The point \((a, a)\) will lie on the line \(y = x\). Depending on the sign of a, it will lie in the first quadrant (if \(a > 0\)) or the third quadrant (if \(a < 0\)), or at the origin (if \(a = 0\)).
In simple words: Points where x and y coordinates are equal lie on the line \(y = x\).

Exam Tip: Always check the signs of coordinates to determine the exact quadrant.

Q12. Where will the point of the form \((a, -a)\) lie?
Answer: The point \((a, -a)\) will lie on the line \(y = -x\). If \(a > 0\), it lies in the fourth quadrant; if \(a < 0\), it lies in the second quadrant.
In simple words: When one coordinate is the negative of the other, the point lies on the line \(y = -x\).

Q13. Write whether True or False. Justify.
(a) \(ax + by + c = 0\) where a, b and c are real numbers, is a linear equation.
(b) A linear equation \(2x + 3y = 5\) has a unique solution.
(c) All the points \((2, 0), (-3, 0), (4, 2)\) and \((0, 5)\) lie on the x-axis.
(d) The line parallel to the y-axis at a distance 4 units to the left of y-axis is given by the equation \(x = -4\).
(e) The graph of the equation \(y = mx + c\) passes through the origin.
Answer:
(a) False. Justification: It must be specified that \(a\) and \(b\) are not both zero (\(a^2 + b^2 \neq 0\)), otherwise it is not a linear equation in two variables.
(b) False. Justification: A linear equation in two variables has infinitely many solutions.
(c) False. Justification: Points \((2, 0)\) and \((-3, 0)\) have y-coordinate zero so they lie on the x-axis, but \((4, 2)\) and \((0, 5)\) do not lie on the x-axis.
(d) True. Justification: A line parallel to the y-axis at a distance 4 units to the left has x-coordinate equal to -4 everywhere, hence \(x = -4\).
(e) False. Justification: It passes through the origin only if \(c = 0\). If \(c \neq 0\), it does not pass through the origin.
In simple words: Check each statement using fundamental definitions of coordinate geometry and linear equations.

Exam Tip: Read every statement carefully for hidden conditions like "a and b are non-zero" or "c equals zero".

Q14. Find the points where the graph of the equation \(3x + 4y = 12\) cuts the x-axis and the y-axis.
Answer:
For x-axis intercept, put \(y = 0\):
\(3x + 4(0) = 12\)
\(3x = 12 \implies x = 4\)
Point on x-axis = \((4, 0)\)
For y-axis intercept, put \(x = 0\):
\(3(0) + 4y = 12\)
\(4y = 12 \implies y = 3\)
Point on y-axis = \((0, 3)\)
In simple words: Set y=0 to find where it cuts the x-axis, and set x=0 to find where it cuts the y-axis.

Exam Tip: Intercept form is a quick way to plot straight lines on a Cartesian plane.

Q15. At what point does the graph \(y = 5\) 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 parallel to the y-axis at a distance of 2 units in the positive direction of the x-axis has the equation \(x = 2\).
The given horizontal line equation is \(y = 5\).
The point of intersection of \(x = 2\) and \(y = 5\) is \((2, 5)\).
In simple words: Combine the x-coordinate from the vertical line and the y-coordinate from the horizontal line.

Exam Tip: The intersection of vertical line \(x = a\) and horizontal line \(y = b\) is always the point \((a, b)\).

Q16. Determine the point on the graph of the equation \(2x + 5y = 20\) whose x-coordinate is \(\frac{2}{5}\) times its ordinate.
Answer:
Let the ordinate (y-coordinate) be y. Then the x-coordinate is \(x = \frac{2}{5}y\).
Substitute \(x = \frac{2}{5}y\) into the equation \(2x + 5y = 20\):
\(2\left(\frac{2}{5}y\right) + 5y = 20\)
\(\frac{4}{5}y + 5y = 20\)
\(\frac{4y + 25y}{5} = 20\)
\(29y = 100 \implies y = \frac{100}{29}\)
Now find x:
\(x = \frac{2}{5}\left(\frac{100}{29}\right) = \frac{40}{29}\)
Therefore, the point is \(\left(\frac{40}{29}, \frac{100}{29}\right)\).
In simple words: Express x in terms of y, substitute into the equation, solve for y, then find x.

Exam Tip: Ordinate means y-coordinate and abscissa means x-coordinate.

Q17. 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:
A line parallel to the x-axis and 4 units above it has the equation \(y = 4\).
To draw its graph, plot points like \((0, 4), (1, 4), (-2, 4)\) on a Cartesian plane and join them with a straight horizontal line.
In simple words: The equation is \(y = 4\), which is a horizontal line passing through 4 on the y-axis.

Exam Tip: For any value of x, y remains fixed at 4.

Q18. Draw the graphs of \(y = x\) and \(y = -x\) on the same cartesian plane.
Answer:
For \(y = x\), points are \((0, 0), (1, 1), (-1, -1)\).
For \(y = -x\), points are \((0, 0), (1, -1), (-1, 1)\).
Both lines pass through the origin \((0, 0)\), forming an 'X' shape intersecting at right angles at the origin.
In simple words: Plot the table of values for both equations and draw lines through them on the same graph.

Exam Tip: Both lines are symmetric with respect to both the x-axis and y-axis.

Q19. If the point \((3, 4)\) lies on the graph \(3y = ax + 7\), then find the value of a.
Answer:
Substitute \(x = 3\) and \(y = 4\) into the equation \(3y = ax + 7\):
\(3(4) = a(3) + 7\)
\(12 = 3a + 7\)
\(3a = 12 - 7\)
\(3a = 5 \implies a = \frac{5}{3}\)
In simple words: Plug the given point coordinates into the equation and solve for the unknown parameter a.

Q20. How many solution(s) of the equation \(2x + 1 = x - 3\) are there on the (i) number line (ii) cartesian plane?
Answer:
First, solve the equation: \(2x + 1 = x - 3 \implies 2x - x = -3 - 1 \implies x = -4\).
(i) On the number line, \(x = -4\) represents a unique single point.
(ii) On the cartesian plane, \(x = -4\) represents a vertical line, which has infinitely many points of the form \((-4, y)\).
In simple words: As an equation in one variable, it has 1 solution; as an equation in two variables, it has infinitely many solutions.

Exam Tip: The interpretation of a solution depends heavily on whether the context is 1D (number line) or 2D (Cartesian plane).

Q21. 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, \(y = 0\):
\(x + 2(0) = 8 \implies x = 8\)
Solution point = \((8, 0)\)
(ii) For a point on the y-axis, \(x = 0\):
\(0 + 2y = 8 \implies y = 4\)
Solution point = \((0, 4)\)
In simple words: Set y to 0 to find the x-axis point, and set x to 0 to find the y-axis point.

Exam Tip: Finding axial intercepts is the standard method for locating where a line touches the axes.

Q22. 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:
Since y varies directly as x, \(y = kx\).
Given \(y = 12\) when \(x = 4\):
\(12 = k(4) \implies k = 3\)
Linear equation: \(y = 3x\)
Now, find y when \(x = 5\):
\(y = 3(5) = 15\)
In simple words: Find the proportionality constant k first, write the equation, and then calculate the required value.

Exam Tip: Direct variation always takes the form \(y = kx\) passing through the origin.

Q23. 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:
Check point A(1, 2):
LHS = \(2\)
RHS = \(9(1) - 7 = 9 - 7 = 2\)
LHS = RHS, so A lies on the graph.
Check point B(-1, -16):
LHS = \(-16\)
RHS = \(9(-1) - 7 = -9 - 7 = -16\)
LHS = RHS, so B lies on the graph.
Check point C(0, -7):
LHS = \(-7\)
RHS = \(9(0) - 7 = 0 - 7 = -7\)
LHS = RHS, so C lies on the graph.
In simple words: Substitute the coordinates of each point into the equation; if both sides match, the point lies on the line.

Exam Tip: All given points must satisfy the equation independently for the statement to be true.

Q24. 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 x-intercept (\(y = 0\)):
\(3x + 4(0) = 6 \implies 3x = 6 \implies x = 2\)
Cuts x-axis at \((2, 0)\).
For y-intercept (\(x = 0\)):
\(3(0) + 4y = 6 \implies 4y = 6 \implies y = \frac{3}{2} = 1.5\)
Cuts y-axis at \(\left(0, \frac{3}{2}\right)\).
Join these two points with a straight line to complete the graph.
In simple words: Calculate where the line crosses both axes, plot those two points, and draw a line through them.

Exam Tip: Two points are sufficient to uniquely determine and draw a straight line graph.

Q25. If the temperature of a liquid can be measured in Kelvin units as \(x^\circ\text{K}\) or Fahrenheit units as \(y^\circ\text{F}\), the relation is given by \(y = \frac{9}{5}(x - 273) + 32\). Draw the graph. (i) Find the temp of the liquid in Fahrenheit if the temp of the liquid is \(313^\circ\text{K}\). (ii) If the temp is \(158^\circ\text{F}\), then find the temp in Kelvin.
Answer:
(i) Given \(x = 313\):
\(y = \frac{9}{5}(313 - 273) + 32\)
\(y = \frac{9}{5}(40) + 32\)
\(y = 9(8) + 32 = 72 + 32 = 104^\circ\text{F}\)
(ii) Given \(y = 158\):
\(158 = \frac{9}{5}(x - 273) + 32\)
\(158 - 32 = \frac{9}{5}(x - 273)\)
\(126 = \frac{9}{5}(x - 273)\)
\(x - 273 = \frac{126 \times 5}{9} = 14 \times 5 = 70\)
\(x = 70 + 273 = 383^\circ\text{K}\)
In simple words: Substitute the given Kelvin or Fahrenheit value into the conversion formula and solve for the unknown temperature.

Exam Tip: Pay careful attention to order of operations when handling fraction multipliers in temperature formulas.

Q26. The linear equation that converts Fahrenheit (F) to Celsius (C) is given by \(C = \frac{5}{9}F - \frac{160}{9}\). Draw the graph.
Answer:
To draw the graph of \(C = \frac{5}{9}F - \frac{160}{9}\):
Choose convenient values for F:
If \(F = 32\), \(C = \frac{5}{9}(32) - \frac{160}{9} = \frac{160 - 160}{9} = 0 \implies (32, 0)\)
If \(F = 212\), \(C = \frac{5}{9}(212) - \frac{160}{9} = \frac{1060 - 160}{9} = \frac{900}{9} = 100 \implies (212, 100)\)
Plot points \((32, 0)\) and \((212, 100)\) on the Cartesian plane and draw a straight line joining them.
In simple words: Pick numbers for F that make calculation clean, find corresponding C values, and plot the line.

Exam Tip: Using freezing point (\(32^\circ\text{F} = 0^\circ\text{C}\)) and boiling point (\(212^\circ\text{F} = 100^\circ\text{C}\)) makes plotting very easy.

Q27. If \(x = 1, y = 2\) is a solution of the equation \(a^2x + ay = 3\), then find the value of a.
Answer:
Substitute \(x = 1\) and \(y = 2\) into the equation \(a^2x + ay = 3\):
\(a^2(1) + a(2) = 3\)
\(a^2 + 2a - 3 = 0\)
Factorize the quadratic equation:
\(a^2 + 3a - a - 3 = 0\)
\(a(a + 3) - 1(a + 3) = 0\)
\((a - 1)(a + 3) = 0\)
Therefore, \(a = 1\) or \(a = -3\).
In simple words: Substitute the coordinates to get a quadratic equation in a, then factorize to find possible values of a.

Exam Tip: Substituting points into equations can sometimes result in quadratic equations yielding multiple valid answers.

Q28. If \(x = 2k - 1\) and \(y = k\) is a solution of the equation \(3x - 5y - 7 = 0\), find the value of k.
Answer:
Substitute \(x = 2k - 1\) and \(y = k\) into the equation \(3x - 5y - 7 = 0\):
\(3(2k - 1) - 5(k) - 7 = 0\)
\(6k - 3 - 5k - 7 = 0\)
\(k - 10 = 0\)
\(k = 10\)
In simple words: Plug the expressions for x and y into the equation and solve the linear equation for k.

Exam Tip: Expand brackets carefully and combine like terms to avoid arithmetic errors.

Q29. If \(x = 2\alpha + 1\) and \(y = \alpha - 1\) is a solution of the equation \(2x - 3y + 5 = 0\), find the value of \(\alpha\).
Answer:
Substitute \(x = 2\alpha + 1\) and \(y = \alpha - 1\) into the equation \(2x - 3y + 5 = 0\):
\(2(2\alpha + 1) - 3(\alpha - 1) + 5 = 0\)
\(4\alpha + 2 - 3\alpha + 3 + 5 = 0\)
\(\alpha + 10 = 0\)
\(\alpha = -10\)
In simple words: Substitute the parametric values into the equation and solve for alpha.

Exam Tip: Watch out for negative signs when distributing the multiplier across parentheses.

Q30. Draw a graph of the line \(x - 2y = 3\). From the graph, find the coordinates of the point when (i) \(x = -5\) (ii) \(y = 0\).
Answer:
First, find points to draw the line \(x - 2y = 3\):
If \(x = 3\), \(3 - 2y = 3 \implies 2y = 0 \implies y = 0 \implies (3, 0)\)
If \(x = 1\), \(1 - 2y = 3 \implies 2y = -2 \implies y = -1 \implies (1, -1)\)
Plot \((3, 0)\) and \((1, -1)\) and draw the line.
(i) When \(x = -5\):
\(-5 - 2y = 3 \implies -2y = 8 \implies y = -4\)
Coordinates = \((-5, -4)\)
(ii) When \(y = 0\):
\(x - 2(0) = 3 \implies x = 3\)
Coordinates = \((3, 0)\)
In simple words: Draw the line using table values, then calculate or read the corresponding coordinates for given x or y values.

Exam Tip: You can verify graphical readings algebraically by substituting values back into the equation.

Q31. Draw the graphs of the equations \(x - y = 1\) and \(2x + y = 8\). Shade the area bounded by these two lines and y-axis.
Answer:
For \(x - y = 1\):
If \(x = 0, y = -1 \rightarrow (0, -1)\)
If \(x = 1, y = 0 \rightarrow (1, 0)\)
For \(2x + y = 8\):
If \(x = 0, y = 8 \rightarrow (0, 8)\)
If \(x = 4, y = 0 \rightarrow (4, 0)\)
Intersection point of the two lines:
Adding both equations: \((x - y) + (2x + y) = 1 + 8 \implies 3x = 9 \implies x = 3\)
Substitute \(x = 3\) into \(x - y = 1 \implies 3 - y = 1 \implies y = 2\)
Intersection point = \((3, 2)\).
The area bounded by the lines and y-axis is the triangle formed by vertices on the y-axis \((0, -1), (0, 8)\) and intersection \((3, 2)\).
In simple words: Plot both lines, find where they intersect each other and the y-axis, and shade the enclosed triangular region.

Exam Tip: Base of the triangle lies along the y-axis, and its length can be found by subtracting y-intercepts (\(8 - (-1) = 9\) units).

Q32. Check whether the point \((a, -a)\) lies on \(x = y + a\) or not.
Answer:
Given equation: \(x = y + a\)
Substitute point coordinates \(x = a\) and \(y = -a\):
LHS = \(a\)
RHS = \(-a + a = 0\)
Since \(a \neq 0\) in general, LHS \(\neq\) RHS.
Therefore, the point \((a, -a)\) does not lie on the line \(x = y + a\).
In simple words: Substitute x and y into the equation and check if both sides match.

Exam Tip: Unless stated \(a = 0\), algebraic substitution must be tested strictly for general non-zero variables.

Q33. Check whether the graph of the equation \(y = 3x + 5\) passes through the origin or not.
Answer:
The origin has coordinates \((0, 0)\).
Substitute \(x = 0\) and \(y = 0\) into \(y = 3x + 5\):
LHS = \(0\)
RHS = \(3(0) + 5 = 5\)
Since \(0 \neq 5\), the point \((0, 0)\) does not satisfy the equation.
Therefore, the graph does not pass through the origin.
In simple words: A line passes through the origin only if its constant term is zero. Here the constant is 5, so it does not pass through the origin.

Exam Tip: Linear equations of the form \(y = mx + c\) pass through the origin only when \(c = 0\).

Q34. Give the equations of two lines passing through \((4, -2)\). How many more such lines are there and why?
Answer:
Two possible linear equations passing through \((4, -2)\):
1. \(x + y = 4 + (-2) \implies x + y = 2\)
2. \(2x - y = 2(4) - (-2) \implies 2x - y = 10\)
There are infinitely many such lines passing through a single point because an infinite number of lines can pass through any given point in a Cartesian plane.
In simple words: You can create infinite equations by combining coordinates, and infinitely many lines pass through one point.

Q35. Solve the equation \(2x + 1 = x - 3\), and represent the solution(s) on (i) the number line (ii) the cartesian plane.
Answer:
Solving the equation:
\(2x + 1 = x - 3\)
\(2x - x = -3 - 1\)
\(x = -4\)
(i) On the number line, the solution \(x = -4\) is represented as a single marked point at -4.
(ii) On the cartesian plane, the solution is represented by the vertical line equation \(x = -4\), which is a line parallel to the y-axis passing through \(x = -4\).
In simple words: Solve for x first to get -4, then show it as a point on a number line or as a vertical line on a graph.

Exam Tip: Always distinguish between single-variable number line representation and two-variable cartesian plane representation.