Latur School Exam Papers 2026
Raja Narayanlal Lahoti English School
Class 9 | Term Test-I
35 Graphing Linear Equations (Internal Choice)
Draw the graphs of the equations \(y=x+3\), \(y=2x+5\) and \(y=3x-2\) and check where these lines cut the y-axis.[cite: 18, 28, 38]
OR
The graph of a linear polynomial \(p(x)\) passes through the points \((1, 5)\) and \((3, 11)\).
(i) Find the polynomial \(p(x)\).
(ii) Find the coordinates where the graph of \(p(x)\) cuts the axes.
(iii) Draw the graph of \(p(x)\).[cite: 18, 28, 38]
(i) Find the polynomial \(p(x)\).
(ii) Find the coordinates where the graph of \(p(x)\) cuts the axes.
(iii) Draw the graph of \(p(x)\).[cite: 18, 28, 38]
Concept:
1
Main: The y-intercept is the point where a line crosses the y-axis. At this specific point, the value of the x-coordinate is always zero. Setting \(x=0\) in any equation \(y = mx+c\) will instantly give the y-intercept.
2
OR: A linear polynomial takes the standard form \(p(x) = ax + b\). When given two coordinate points \((x, y)\), you can set up a system of simultaneous equations to find the constants 'a' and 'b'.
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Hint:
1
Main: To draw the graphs, pick 2 easy x-values (like 0 and 1) for each equation to find the corresponding y-values. Plot the points and draw the straight lines. To find where they cut the y-axis, simply plug in \(x=0\).
2
OR: Substitute \(x=1, p(x)=5\) to get \(a + b = 5\). Substitute \(x=3, p(x)=11\) to get \(3a + b = 11\). Solve these two equations. Then, set \(x=0\) to find the y-axis cut, and \(p(x)=0\) to find the x-axis cut.
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Mistakes to Avoid:
!
Axis Confusion: Mixing up the x and y intercepts. The line cuts the y-axis when \(x = 0\). The line cuts the x-axis when \(y = 0\). Writing \((3, 0)\) instead of \((0, 3)\) for a y-intercept is a fatal error.
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MAIN QUESTION SOLUTION:
A line cuts the y-axis where the x-coordinate is 0.
1. For \( y = x + 3 \), put \( x = 0 \implies y = 3 \). Point: \( (0, 3) \) [½ M]
2. For \( y = 2x + 5 \), put \( x = 0 \implies y = 5 \). Point: \( (0, 5) \) [½ M]
3. For \( y = 3x - 2 \), put \( x = 0 \implies y = -2 \). Point: \( (0, -2) \) [½ M]
[3.5 M for Combined Graphs]
OR
OR QUESTION SOLUTION:
Let the linear polynomial be \( p(x) = ax + b \).
Passes through \( (1, 5) \implies a(1) + b = 5 \implies a + b = 5 \) --- (Eq 1)
Passes through \( (3, 11) \implies a(3) + b = 11 \implies 3a + b = 11 \) --- (Eq 2)
Subtracting Eq 1 from Eq 2:
\( (3a + b) - (a + b) = 11 - 5 \implies 2a = 6 \implies a = 3 \).
Substitute \( a = 3 \) in Eq 1: \( 3 + b = 5 \implies b = 2 \).
(i) Polynomial: \( p(x) = 3x + 2 \) [1.5 M]
(ii) Intercepts:
Cuts y-axis when \( x=0 \): \( p(0) = 3(0) + 2 = 2 \). Point: \( (0, 2) \)
Cuts x-axis when \( p(x)=0 \): \( 3x + 2 = 0 \implies x = -2/3 \). Point: \( (-2/3, 0) \) [1.5 M]
(iii) Graph: The line is plotted above passing through (1,5) and (3,11). [2 M]
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