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Maths 4 Marks Questions

Linear Equations In Two Variables · CBSE STD 9 (CBSE - English Medium). 15 questions.

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4 Marks Questions

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15 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks
Draw the graph of $y = |x|$.
Answer
We have,
$y = |X| ...(i)$
Putting $x = 0$, we get $y = 0$
Putting $x = 2$, we get $y = 2$
Putting $x = -2$, we get $y = 2$
Thus, we have the following table for the points on graph of $|x|$.
x
$0$
$2$
$-2$
y
$0$
$2$
$2$
The graph of the equation $y = |x|$:

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Question 24 Marks
Draw the graph of the following linear equations in two variables: $x + y = 4$
Answer
We have, $x + y = 4$
$\Rightarrow x = 4 - y ...(i)$
Putting $y = 0$,
we get $x = 4 - 0 = 4$
Putting $y = 3,$
we get $x = 4 - 3 = 1$
Thus, we have the following table giving two points on the line represented by the equation $x + y = 4$:
Graph of the equation $x + y = 4$:
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Question 34 Marks
Draw the graph of the following linear equations in two variables: $y = 2x$
Answer
We have, $y = 2x ...(i)$
Putting $x = 0$, we get $y = 2 \times 0 = 0$
Putting $x = 1$, we get $y = 2 \times 1 = 2$
Thus, we have the following table giving two points on the line represented by the equation $y = 2x$:
Graph of the equation $y = 2x$:
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Question 44 Marks
Draw the graph of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes: $3x + 2y + 6 = 0$
Answer
we have, $3x + 2y + 6 = 0$
 $\Rightarrow 2y = -6 - 3x$
$\Rightarrow\text{y}=\frac{-6-3\text{x}}{2}\ ...(\text{i})$
Putting $x = -2$ in $(i)$, we get $\text{y}=\frac{6-3(-2)}{2}=0$
Putting $x = -4$ in $(i)$, we get $\text{y}=\frac{-6-3(-4)}{2}=3$
Thus, we obtain the following table giving coordinates of two points on the line represented by the equation $3x + 2y + 6 = 0$.
The graph of line $3x + 2y + 6 = 0$:
 
Clearly, the line intersect with the coordinate axes $(-2, 0)$ and $(0, -3)$.
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Question 54 Marks
Draw the graph of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes: $6x - 3y = 12$
Answer
We have, $6x - 3y = 12 $
$\Rightarrow 3(2x - y) = 12 $
$\Rightarrow 2x - y = 4 $
$\Rightarrow 2x - 4 = y $
$\Rightarrow y = 2x - 4 ...(i)$
Putting $x = 0$ in $(i)$,
we get $y = -4$
Putting $x = 2$ in $(i)$,
we get $y = 0$
Thus, we obtain the following table giving coordinates of two points on the line represented by the equation $6x - 3y = 12$
The graph of line $6x - 3y = 12$:
 
Clearly, the line intersect with the coordinate axes $(2, 0)$ and $(0, -4)$.
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Question 64 Marks
Draw the graph of the following linear equations in two variables: $2y = -x + 1$
Answer
We have, $2y = -x + 1$
$\Rightarrow x = 1 - 2y ...(i)$
Putting $y = 0$, we get $x = 1 - 2 \times 0 = 1$
Putting $y = -1$ we get $x = 1 - 2(-1) = 3$
Thus, we have the following table giving two points on the line represented by the equation $2y = -x + 1$:
Graph of the equation $2y = -x + 1$:
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Question 74 Marks
Draw the graph of $y = |x| + 2$.
Answer
We have, $y = |x| + 2 ...(i)$
Putting $x = 0$,
we get $y = 2$ Putting $x = 1$,
we get $y = 3$ Putting $x = -1$,
we get $y = 3$
Thus, we have the following table for the points on graph of $|x| + 2$:
$x$
$0$
$1$
$-1$
$y$
$2$
$3$
$3$
The graph of the equation $y = |x| + 2$:
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Question 84 Marks
Draw the graph of the following linear equations in two variables: $x - y = 2$
Answer
We have, $x - y = 2$ $\Rightarrow x = 2 + y ...(i)$
Putting $y = 0$, we get $x = 2 + 0 = 2$
Putting $y = -2$, we get $x = 2 - 2 = 0$
Thus, we have the following table giving two points on the line represented by the equation $x - y = 2$: Graph of the equation $x - y = 2$:
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Question 94 Marks
Draw the graph of the following linear equations in two variables: $\frac{\text{x}-2}{3}=\text{y}-3$
Answer
We have, $\frac{\text{x}-2}{3}=\text{y}-3$
$\Rightarrow x - 2 = 3(y - 3)$
$\Rightarrow x - 2 = 3y - 9$
$\Rightarrow x = 3y - 9 + 2$
$\Rightarrow x = 3y - 7 ...(i)$
Putting $y = 2$, we get $x = 3(2) - 7 = -1$
Putting $y = 3$, we get $x = 3(3) - 7 = 2$
Thus, we have the following table giving two points on the line represented by the equation $\frac{\text{x}-2}{3}=\text{y}-3:$
Graph of the equation $\frac{\text{x}-2}{3}=\text{y}-3:$
​​​​​​​
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Question 104 Marks
Draw the graph of the following linear equations in two variables: $3x + 5y = 15$
Answer
We have, $3x + 5y = 15$
$\Rightarrow 3x = 15 - 5y$
$\Rightarrow\text{x}=\frac{15-5\text{y}}{3}\ ...\text{(i)}$
Putting $y = 0$, we get $\text{x}=\frac{15-5\times0}{3}=5$
Putting $y = 3$, we get $\text{x}=\frac{15-5\times3}{3}=0$
Thus, we have the following table giving two points on the line represented by the equation $3x + 5y = 15$:
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Question 114 Marks
Draw the graphs of the following linear equations on the same graph paper: $2x + 3y = 12, x - y = 1$ Find the coordinates of the vertices of the triangle formed by the two straight lines and the $y$-axis. Also, find the area of the triangle.
Answer
Graph of the equation $2x + 3y = 12$:
We have,
$2x + 3y = 12$
$\Rightarrow 2x = 12 - 3y$
$\Rightarrow\text{x}=\frac{12-3\text{y}}{2}\ ...\text{(i)}$
Putting $y = 4$, we get $\text{x}=\frac{12-3\times4}{2}=0$
Putting $y = 2$, we get $\text{x}=\frac{12-3\times2}{2}=3$
Thus, we have the following table for the points on the line $2x + 3y = 12$:
$x$ $0$ $3$
$y$ $4$ $2$
Ploting points $A(0, 4), B(3, 2)$ on the graph paper and drawing a line passing through them, we obtain graph of the equation $2x + 3y = 12$.
Graph of the equation $x - y = 1$:
We have,
$x - y = 1$
$\Rightarrow x = 1 + y$
Putting $y = 0$, we get $x = 1 + 0 = 1$
Putting $y = -1$, we get $x = 1 - 1 = 0$
Thus, we have the following table for the points on the line $x - y = 1$:
$x$ $1$ $0$
$y$ $0$ $-1$
Ploting points $C(1, 0)$ and $D(0, -1)$ on the same graph paper and drawing a line passing through them, we obtain the graph of the line represented by the equation $x - y = 1$.

Clearly, two lines intersect at $A(3, 2)$.
The graph of line $2x + 3y = 12$ intersect with y-axis at $B(0, 4)$ and the graph of the line $x - y = 1$ intersect with $y$-axis at $C(0, -1)$.
So, the vertices of the triangle formed by the two straight lines and $y$-axis are $A(3, 2), B(0, 4)$ and $C(0, -1)$.
Now,
$\text{Area of }\triangle\text{ABC}=\frac{1}{2}(\text{Base}\times\text{Height})$
$=\frac{1}{2}(\text{BC}\times\text{AM})$
$=\frac{1}{2}(5\times3)$
$=\frac{15}{2}\text{sq.units.}$
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Question 124 Marks
Draw the graph of the following linear equations in two variables: $\frac{\text{x}}{2}-\frac{\text{y}}{3}=2$
Answer
We have, $\frac{\text{x}}{2}-\frac{\text{y}}{3}=2$
$\Rightarrow\frac{3\text{x}-2\text{y}}{6}=2$
$\Rightarrow3\text{x}-2\text{y}=12$
$\Rightarrow3\text{x}=12+2\text{y}$
$\Rightarrow\text{x}=\frac{12+2\text{y}}{3}$
Putting $y = -6$, we get $\text{x}=\frac{12+2(-6)}{3}=0$
Putting $y = -3$, we get $\text{x}=\frac{12+2(-3)}{3}=2$
Thus, we have the following table giving two points on the line represented by the equation
$\frac{\text{x}}{2}-\frac{\text{y}}{3}=2:$
Graph of the equation $\frac{\text{x}}{2}-\frac{\text{y}}{3}=2:$
 
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Question 134 Marks
Draw the graph of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes: $2x + y = 6$
Answer
we have, $2x + y = 6 \Rightarrow y = 6 - 2x ...(i)$ Putting $x = 3$ in $(i)$,
we get $y = 6 - 2 × 3 = 0$ Putting $x = 4$ in $(i)$,
we get $y = 6 - 2 × 4 = -2$ Thus, we obtain the following table giving coordinates of two points on the line represented by the equation $2x + y = 6$.
The graph of line $2x + y = 6$:

Clearly, the line intersect with the coordinate axes $(3, 0)$ and $(0, 6)$.
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Question 144 Marks
Draw the graph of the following linear equations in two variables:$ -x + y = 6$
Answer
We have, $-x + y = 6$
$\Rightarrow y = 6 + x ...(i)$
Putting $x = -4,$ we get $y = 6 - 4 = 2$
Putting $x = -3,$ we get $y = 6 - 3 = 3$
Thus, we have the following table giving two points on the line represented by the equation $-x + y = 6$
Graph of the equation $-x + y = 6$
:
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Question 154 Marks
Draw the graph of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes:$ -x + 4y = 8$
Answer
we have,$ -x + 4y = 8 \Rightarrow 4y - 8 = x \Rightarrow x = 4y - 8 ...(i)$
Putting $y = 1$ in $(i),$ we get $x = 4 × 1 - 8 = -4$
Putting $y = 2$ in $(i),$ we get $x = 4 × 2 - 8 = 0$
Thus, we obtain the following table giving coordinates of two points on the line represented by the equation $-x + 4y = 8$.
The graph of line $-x + 4y = 8$:
 
Clearly, the line intersect with the coordinate axes $(-8, 0)$ and $(0, 2)$.
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4 Marks Questions - Maths STD 9 Questions - Vidyadip