Welcome to Sid Classes. In this extensive guide, we provide accurate, step-by-step solutions for NCERT Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.6. This graphing exercise helps students visually understand how the variables \(a\) (slope) and \(b\) (\(y\)-intercept) control the steepness, direction, and placement of a line on a Cartesian plane.
Our solutions use clean standard mathematical notation (\(\text{LaTeX}\)) and breakdown logic to ensure you score 100% marks in your school tests and board examinations.
🔑 Key Concepts to Remember:
- The Role of \(a\) (Slope): Determines the steepness and direction of the line. If \(a > 0\), the line slants upwards from left to right. If \(a < 0\), the line slants downwards. A larger absolute value of \(a\) creates a steeper line.
- The Role of \(b\) (\(y\)-intercept): Determines where the line crosses the vertical \(y\)-axis. It shifts the line vertically up (if \(b > 0\)) or down (if \(b < 0\)). If \(b = 0\), the line passes directly through the origin \((0,0)\).
NCERT Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.6 Solutions
Question 1 (i)
Draw the graphs of the set of lines: \(y = 4x\), \(y = 2x\), \(y = x\). Reflect on the role of ‘a’ and ‘b’.
Solution:
To plot these lines, let's find the coordinate points by substituting values for \(x\):
- For \(y = 4x\): \((-1, -4)\), \((0, 0)\), \((1, 4)\)
- For \(y = 2x\): \((-1, -2)\), \((0, 0)\), \((1, 2)\)
- For \(y = x\): \((-1, -1)\), \((0, 0)\), \((1, 1)\)
Reflection on ‘a’ and ‘b’:
Here, the values of \(a\) are \(4\), \(2\), and \(1\), while \(b = 0\) for all lines.
* Role of \(b\): Since \(b = 0\), all three lines pass exactly through the origin \((0,0)\).
* Role of \(a\): All values of \(a\) are positive, so all lines tilt upwards. As \(a\) increases from \(1 \rightarrow 2 \rightarrow 4\), the line rotates counterclockwise and becomes progressively steeper towards the vertical \(y\)-axis.
Question 1 (ii)
Draw the graphs of the set of lines: \(y = -6x\), \(y = -3x\), \(y = -x\). Reflect on the role of ‘a’ and ‘b’.
Solution:
Let's calculate key coordinate values to construct the graph plane points:
- For \(y = -6x\): \((-1, 6)\), \((0, 0)\), \((1, -6)\)
- For \(y = -3x\): \((-1, 3)\), \((0, 0)\), \((1, -3)\)
- For \(y = -x\): \((-1, 1)\), \((0, 0)\), \((1, -1)\)
Reflection on ‘a’ and ‘b’:
Here, the values of \(a\) are \(-6\), \(-3\), and \(-1\), while \(b = 0\).
* Role of \(b\): Because \(b = 0\), every line passes directly through the origin \((0,0)\).
* Role of \(a\): All values of \(a\) are negative, meaning all lines slope downwards from left to right. As the value gets more negative (\(-1 \rightarrow -3 \rightarrow -6\)), the absolute magnitude grows, causing the lines to get increasingly steeper.
Question 1 (iii)
Draw the graphs of the set of lines: \(y = 5x\), \(y = -5x\). Reflect on the role of ‘a’ and ‘b’.
Solution:
Let's find the plotting coordinates for comparison:
- For \(y = 5x\): \((-1, -5)\), \((0, 0)\), \((1, 5)\)
- For \(y = -5x\): \((-1, 5)\), \((0, 0)\), \((1, -5)\)
Reflection on ‘a’ and ‘b’:
* Role of \(b\): Both lines have \(b = 0\), meaning they intersect at the origin \((0,0)\).
* Role of \(a\): The slopes have the exact same absolute value (\(|5| = |-5| = 5\)), meaning they share identical steepness. However, because one is positive (\(+5\)) and one is negative (\(-5\)), they are perfect mirror reflections of each other across the \(y\)-axis.
Question 1 (iv)
Draw the graphs of the set of lines: \(y = 3x - 1\), \(y = 3x\), \(y = 3x + 1\). Reflect on the role of ‘a’ and ‘b’.
Solution:
Let's find key coordinate point pairings for these lines:
- For \(y = 3x - 1\): \((0, -1)\), \((1, 2)\), \((2, 5)\)
- For \(y = 3x\): \((0, 0)\), \((1, 3)\), \((2, 6)\)
- For \(y = 3x + 1\): \((0, 1)\), \((1, 4)\), \((2, 7)\)
Reflection on ‘a’ and ‘b’:
* Role of \(a\): The value of \(a = 3\) is identical across all three equations. Since their slopes are equal, these lines climb at the exact same rate, making them perfectly parallel lines that never intersect.
* Role of \(b\): The \(b\) values are \(-1\), \(0\), and \(+1\). This shifts the lines vertically. The line \(y = 3x+1\) is shifted up by 1 unit from the origin, while \(y = 3x-1\) is shifted downward by 1 unit.
Question 1 (v)
Draw the graphs of the set of lines: \(y = -2x - 3\), \(y = -2x\), \(y = 2x + 3\). Reflect on the role of ‘a’ and ‘b’.
Solution:
Let's find the plotting graph coordinates for this mixed collection set:
- For \(y = -2x - 3\): \((0, -3)\), \((-1, -1)\), \((-2, 1)\)
- For \(y = -2x\): \((0, 0)\), \((1, -2)\), \((-1, 2)\)
- For \(y = 2x + 3\): \((0, 3)\), \((1, 5)\), \((-1, 1)\)
Reflection on ‘a’ and ‘b’:
* Comparing line 1 (\(y = -2x-3\)) and line 2 (\(y = -2x\)): They have the same slope (\(a = -2\)), so they run parallel to each other. Line 1 is shifted downward by 3 units because its \(b = -3\).
* Comparing line 1 (\(y = -2x-3\)) and line 3 (\(y = 2x+3\)): Their slopes are opposite (\(-2\) vs \(+2\)), so one runs downwards and the other upwards with equal steepness.
* Comparing line 2 (\(y = -2x\)) and line 3 (\(y = 2x+3\)): Line 2 passes through the origin (\(b=0\)), whereas line 3 intersects the \(y\)-axis higher up at \((0,3)\) because its \(b = +3\).
Frequently Asked Questions (FAQs)
Q1: What happens to a linear graph when 'a' is equal to zero?
A: When \(a = 0\), the equation simplifies to \(y = b\). This creates a perfectly horizontal line parallel to the \(x\)-axis that cuts through the value of \(b\) on the vertical axis.
Q2: How do parallel lines look when written in the form \(y = ax + b\)?
A: Parallel lines will always have the exact same value for \(a\) (slope) but will feature completely different values for the constant \(b\) (\(y\)-intercept).
Q3: What does a negative slope value tell us about a line graph?
A: A negative slope (\(a < 0\)) indicates that the line falls downwards from left to right. As values on the horizontal \(x\)-axis increase, the matching \(y\)-axis coordinates decrease proportionally.
Q4: What is the origin on a Cartesian plane?
A: The origin is the point where the horizontal \(x\)-axis and vertical \(y\)-axis cross each other, located precisely at the coordinates \((0,0)\).
Q5: Can we safely copy this HTML layout into custom Blogger template formats?
A: Yes. Because this layout avoids strict background overrides and centers on universal inline padding styles, it automatically renders perfectly across all custom blogging templates.
Q6: Are these structural layouts updated for the latest NCERT Ganita Manjari Grade 9 editions?
A: Yes, these solution parameters perfectly track the design concepts mandated under Chapter 2 patterns inside the contemporary Part I textbook guidelines.
Quick Navigation:
Last updated according to the latest NCERT Ganita Manjari syllabus.