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#### Question 1:

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case.
(i) 3x + 5y = 7.5
(ii) $2x-\frac{y}{5}+6=0$
(iii) 3y – 2x = 6
(iv) 4x = 5y
(v) $\frac{x}{5}-\frac{y}{6}=1$
(vi) $\sqrt{2}x+\sqrt{3}y=5$

(i) 3x + 5= 7.5
This can be expressed in the form ax + by + c = 0 as $3x+5y+\left(-7.5\right)=0$.
(ii) $2x-\frac{y}{5}+6=0$
This can be expressed in the form ax + by + c = 0 as $2x+\left(-\frac{1}{5}\right)y+6=0$
(iii) 3y – 2x = 6
This can be expressed in the form ax + by + c = 0 as $2x+\left(-3y\right)+6=0$.
(iv) 4x = 5y
This can be expressed in the form ax + by + c = 0 as $4x-5y+0=0$
(v) $\frac{x}{5}-\frac{y}{6}=1$
This can be expressed in the form ax + by + c = 0 as $6x-5y=30$
(vi) $\sqrt{2}x+\sqrt{3}y=5$
This can be expressed in the form ax + by + c = 0 as $\sqrt{2}x+\sqrt{3}y+\left(-5\right)=0$

#### Question 2:

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case.
(i) x = 6
(ii) 3xy = x – 1
(iii) 2x + 9 = 0
(iv) 4y = 7
(v) x + y = 4
(vi) $\frac{x}{2}-\frac{y}{3}=\frac{1}{6}+y$

(i) x = 6
In the form of ax + by + c = 0 we have $x+0y+\left(-6\right)=0$ where a = 1, b = 0 and $c=-6$.
(ii) 3x – y = x – 1
In the form of ax + by + c = 0 we have $2x+\left(-1y\right)+1=0$ where a = 2, b = $-1$ and c = 1
(iii) 2x + 9 = 0
In the form of ax + by + c = 0 we have $2x+\left(-1y\right)+1=0$ where a = 2, b = $-1$ and c = 1
(iv) 4y = 7
In the form of ax + by + c = 0 we have $0x+4y+\left(-7\right)=0$ where a = 0, b = 4 and c = $-7$
(v) x + y = 4
In the form of ax + by + c = 0 we have $1x+1y+\left(-4\right)=0$ where $a=1,b=1,c=-4$
(vi) $\frac{x}{2}-\frac{y}{3}=\frac{1}{6}+y$
In the form of ax + by + c = 0 we have $3x-2y=1+6y⇒3x-8y+\left(-1\right)=0\phantom{\rule{0ex}{0ex}}$ where $a=3,b=-8,c=-1$.

#### Question 3:

Check which of the following are the solutions of the equation 5x – 4y = 20.
(i) (4, 0)
(ii) (0, 5)
(iii)
(iv) (0, –5)
(v)

The equation given is 5x – 4y = 20.
(i) (4, 0)
Putting the value in the given equation we have
$\mathrm{LHS}:5\left(4\right)-4\left(0\right)=20\phantom{\rule{0ex}{0ex}}\mathrm{RHS}:20\phantom{\rule{0ex}{0ex}}\mathrm{LHS}=\mathrm{RHS}$
Thus, (4, 0) is a solution of the given equation.
(ii) (0, 5)
Putting the value in the given equation we have
$\mathrm{LHS}:5\left(0\right)-4\left(5\right)=0-20=-20\phantom{\rule{0ex}{0ex}}\mathrm{RHS}:20\phantom{\rule{0ex}{0ex}}\mathrm{LHS}\ne \mathrm{RHS}$
Thus, (0, 5) is not a solution of the given equation.
(iii)
Putting the value in the given equation we have
$\mathrm{LHS}:5\left(-2\right)-4\left(\frac{5}{2}\right)=-10-10=-20\phantom{\rule{0ex}{0ex}}\mathrm{RHS}:20\phantom{\rule{0ex}{0ex}}\mathrm{LHS}\ne \mathrm{RHS}$
Thus, is not a solution of the given equation.
(iv) (0, –5)
Putting the value in the given equation we have
$\mathrm{LHS}:5\left(0\right)-4\left(-5\right)=0+20=20\phantom{\rule{0ex}{0ex}}\mathrm{RHS}:20\phantom{\rule{0ex}{0ex}}\mathrm{LHS}=\mathrm{RHS}$
Thus, (0, –5) is a solution of the given equation.
(v)
Putting the value in the given equation we have
$\mathrm{LHS}:5\left(2\right)-4\left(\frac{-5}{2}\right)=10+10=20\phantom{\rule{0ex}{0ex}}\mathrm{RHS}:20\phantom{\rule{0ex}{0ex}}\mathrm{LHS}=\mathrm{RHS}$
Thus,  is a solution of the given equation.

#### Question 4:

Find five different solutions of each of the following equations:
(i) 2x – 3y = 6
(ii) $\frac{2x}{5}+\frac{3y}{10}=3$
(iii) 3y = 4x

(i) 2x – 3y = 6

 x 0 3 $-$3 $\frac{9}{2}$ 2 y $-$2 0 $-$4 1 $\frac{-2}{3}$

(ii) $\frac{2x}{5}+\frac{3y}{10}=3$
 x 0 $\frac{15}{2}$ 5 10 3 y 10 0 $\frac{10}{3}$ $\frac{-10}{3}$ 6

(iii) 3y = 4x

 x 3 $-$3 $-$6 6 0 y 4 $-$4 $-$8 8 0

#### Question 5:

If x = 3 and y = 4 is a solution of the equation 5x – 3y = k, find the value of k.

Given: 5x – 3y = k
Since x = 3 and = 4 is a solution of the given equation so, it should satisfy the equation.
$5\left(3\right)-3\left(4\right)=k\phantom{\rule{0ex}{0ex}}⇒15-12=k\phantom{\rule{0ex}{0ex}}⇒3=k$

#### Question 6:

If x = 3k + 2 and y = 2k – 1 is a solution of the equation 4x – 3y + 1 = 0, find the value of k.

Given: 4– 3+ 1 = 0                                 .....(1)
x = 3k + 2 and y = 2k – 1
Putting these values in the equation (1) we get
$4\left(3k+2\right)-3\left(2k-1\right)+1=0\phantom{\rule{0ex}{0ex}}⇒12k+8-6k+3+1=0\phantom{\rule{0ex}{0ex}}⇒6k+12=0\phantom{\rule{0ex}{0ex}}⇒k+2=0\phantom{\rule{0ex}{0ex}}⇒k=-2$

#### Question 7:

The cost of 5 pencils is equal to the cost of 2 ballpoints. Write a linear equation in two variables to represent this statement. (Take the cost of a pencil to be ₹ x and that of a ballpoint to be ₹ y).

Let cost of a pencil to be ₹ x and that of a ballpoint to be ₹ y.
Cost of 5 pencils = 5x
Cost of 2 ballpoints = 2y
Cost of 5 pencils = cost of 2 ballpoints
$⇒5x=2y\phantom{\rule{0ex}{0ex}}⇒5x-2y=0$

#### Question 1:

Draw the graph of each of the following equations.
(i) x = 4
(ii) x + 4 = 0
(iii) y = 3
(iv) y = –3
(v) x = –2
(vi) x = 5
(vii) y + 5 = 0
(viii) y = 4

(i) The equation of given line is x = 4. This equation does not contain the term of y. So, the graph of this line is parallel to y-axis passing through the point (4, 0).

(ii) The equation of given line is x + 4 = 0 or x = –4. This equation does not contain the term of y. So, the graph of this line is parallel to y-axis passing through the point (–4, 0).

(iii) The equation of given line is y = 3. This equation does not contain the term of x. So, the graph of this line is parallel to x-axis passing through the point (0, 3).

(iv) The equation of given line is y = –3. This equation does not contain the term of x. So, the graph of this line is parallel to x-axis passing through the point (0, –3).

(v) The equation of given line is x = –2. This equation does not contain the term of y. So, the graph of this line is parallel to y-axis passing through the point (–2, 0).

(vi) The equation of given line is x = 5. This equation does not contain the term of y. So, the graph of this line is parallel to y-axis passing through the point (5, 0).

(vii) The equation of given line is y + 5 = 0 or y = –5. This equation does not contain the term of x. So, the graph of this line is parallel to x-axis passing through the point (0, –5).

(viii) The equation of given line is y = 4. This equation does not contain the term of x. So, the graph of this line is parallel to x-axis passing through the point (0, 4).

#### Question 2:

Draw the graph of the equation y = 3x.
From your graph, find the value of y when
(i) x = 2
(ii) x = –2.

Given equation:  y = 3x

When x = -2, y = -6.
When x = -1, y = -3.
Thus, we have the following table:

 x -2 -1 y -6 -3

Now plot the points (-2,-6), (-1, -3) on a graph paper.
Join the points and extend the line in both the directions.
The line segment is the required  graph of the equation.
From the graph we can see that when x = 2, y = 6

Also, when x = -2, y = -6.

#### Question 3:

Draw the graph of the equation x + 2y – 3 = 0.
From your graph, find the value of y when
(i) x = 5
(ii) x = –5.

Given equation: x + 2y - 3 = 0
Or,  x + 2y = 3
When y = 0, x + 0 = 3 ⇒ x = 3
When y = 1, x + 2 = 3 ⇒ x = 3-2 = 1
When y = 2, x + 4 = 3 ⇒ x = 3 - 4 = -1
Thus, we have the following table:

 x 3 1 -1 y 0 1 2

Now plot the points (3,0) ,(1,1) and (-1,2) on the graph paper.
Join the points and extend the line in both the directions.
The line segment is the required graph of the equation.

When x = 5,

Similarly, from the graph we can see that when x = −5, y = 4.

#### Question 4:

Draw the graph of the equation 2x − 3y = 5. From the graph, find (i) the value of y when x = 4 and (ii) the value of x when y = 3.

Given equation :

When,

When,
Thus, we have the following table:

 $\text{x}$ 1 -2 $y$ -1 -3
Plot the points  on the graph paper and extend the line in both directions.

(i) When x = 4:

(ii) When y = 3:

#### Question 5:

Draw the graph of the equation 2x + y = 6. Find the coordinates of the point, where the graph cuts the x-axis.

Given equation:

When, .
When, .
When, .
Thus, we have the following table:

 $\text{x}$ 0 1 2 $y$ 6 4 2

Plot the points  and $\left(2,2\right)$ on the graph paper. Join these points and extend the line.

Clearly, the graph cuts the $x-axis$ at $P\left(3,0\right)$.

#### Question 6:

Draw the graph of the equation 3x + 2y = 6. Find the coordinates of the point, where the graph cuts the y-axis.

Given equation: . Then,

When ,

When ,
Thus, we get the following table:

 $\text{x}$ 2 4 $y$ 0 -3

Plot the points  on the graph paper. Join the points and extend the graph in both the directions.

Clearly, the graph cuts the $y-axis$ at P(0,3).

#### Question 7:

Draw the graphs of the equations 3x – 2y = 4 and x + y – 3 = 0. On the same graph paper, find the coordinates of the point were the two graph lines intersect.

$3x-2y=4\phantom{\rule{0ex}{0ex}}⇒2y=3x-4\phantom{\rule{0ex}{0ex}}⇒y=\frac{3x-4}{2}$
When x = 0, $y=\frac{3×0-4}{2}=\frac{0-4}{2}=\frac{-4}{2}=-2$
When x = 2, $y=\frac{3×2-4}{2}=\frac{6-4}{2}=\frac{2}{2}=1$
When x = –2, $y=\frac{3×\left(-2\right)-4}{2}=\frac{-6-4}{2}=\frac{-10}{2}=-5$
Thus, the points on the line 3x – 2y = 4 are as given in the following table:

 x 0 2 –2 y –2 1 –5

Plotting the points (0, –2), (2, 1) and (–2–5) and drawing a line passing through these points, we obtain the graph of of the line 3x – 2y = 4.

$x+y-3=0\phantom{\rule{0ex}{0ex}}⇒y=-x+3$
When x = 0, $y=-0+3=3$
When x = 1, $y=-1+3=2$
When x = –1, $y=-\left(-1\right)+3=1+3=4$
Thus, the points on the line x + y – 3 = 0 are as given in the following table:
 x 0 1 –1 y 3 2 4

Plotting the points (0, 3), (1, 2) and (–14) and drawing a line passing through these points, we obtain the graph of of the line x + y – 3 = 0.

It can be seen that the lines 3x – 2y = 4 and x + y – 3 = 0 intersect at the point (2, 1).

#### Question 8:

Draw the graph of the line 4x + 3y = 24.
(i) Write the coordinates of the points where this line intersects the x-axis and the y-axis.
(ii) Use this graph to find the area of the triangle formed by the graph line and the coordinate axes.

$4x+3y=24\phantom{\rule{0ex}{0ex}}⇒3y=-4x+24\phantom{\rule{0ex}{0ex}}⇒y=\frac{-4x+24}{3}$
When x = 0, $y=\frac{-4×0+24}{3}=\frac{0+24}{3}=\frac{24}{3}=8$
When x = 3, $y=\frac{-4×3+24}{3}=\frac{-12+24}{3}=\frac{12}{3}=4$
When x = 6, $y=\frac{-4×6+24}{3}=\frac{-24+24}{3}=\frac{0}{3}=0$
Thus, the points on the line 4x + 3y = 24 are as given in the following table:

 x 0 3 6 y 8 4 0

Plotting the points (0, 8), (3, 4) and (60) and drawing a line passing through these points, we obtain the graph of of the line 4x + 3y = 24.

(i) It can be seen that the line 4x + 3y = 24 intersects the x-axis at (6, 0) and y-axis at (0, 8).

(ii) The triangle formed by the line and the coordinate axes is a right triangle right angled at the origin.

∴ Area of the triangle = $\frac{1}{2}×6×8$ = 24 square units

#### Question 9:

Draw the graphs of the lines 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by these lines and the x-axis. Find the area of the shaded region.

$2x+y=6\phantom{\rule{0ex}{0ex}}⇒y=-2x+6$
When x = 0, $y=-2×0+6=0+6=6$
When x = 1, $y=-2×1+6=-2+6=4$
When x = 2, $y=-2×2+6=-4+6=2$

Thus, the points on the line 2x + y = 6 are as given in the following table:

 x 0 1 2 y 6 4 2

Plotting the points (0, 6), (1, 4) and (22) and drawing a line passing through these points, we obtain the graph of of the line 2x + y = 6.

$2x-y+2=0\phantom{\rule{0ex}{0ex}}⇒y=2x+2$
When x = 0, $y=2×0+2=0+2=2$
When x = 1, $y=2×1+2=2+2=4$
When x = –1, $y=2×\left(-1\right)+2=-2+2=0$

Thus, the points on the line 2x – y + 2 = 0 are as given in the following table:
 x 0 1 –1 y 2 4 0

Plotting the points (0, 2), (1, 4) and (–10) and drawing a line passing through these points, we obtain the graph of of the line 2x – y + 2 = 0.

The shaded region represents the area bounded by the lines 2x + y = 6, 2x – y + 2 = 0 and the x-axis. This represents a triangle.

It can be seen that the lines intersect at the point C(1, 4). Draw CD perpendicular from C on the x-axis.

Height = CD = 4 units

Base = AB = 4 units

∴ Area of the shaded region = Area of ∆ABC = $\frac{1}{2}×\mathrm{AB}×\mathrm{CD}=\frac{1}{2}×4×4$ = 8 square units

#### Question 10:

Draw the graphs of the lines x – y = 1 and 2x + y = 8. Shade the area formed by these two lines and the y-axis. Also, find this area.

$x-y=1\phantom{\rule{0ex}{0ex}}⇒y=x-1$
When x = 0, $y=0-1=-1$
When x = 1, $y=1-1=0$
When x = 2, $y=2-1=1$

Thus, the points on the line x – y = 1 are as given in the following table:

 x 0 1 2 y –1 0 1

Plotting the points (0, –1), (1, 0) and (21) and drawing a line passing through these points, we obtain the graph of of the line x – y = 1.

$2x+y=8\phantom{\rule{0ex}{0ex}}⇒y=-2x+8$
When x = 1, $y=-2×1+8=-2+8=6$
When x = 2, $y=-2×2+8=-4+8=4$
When x = 3, $y=-2×3+8=-6+8=2$

Thus, the points on the line  2x + y = 8 are as given in the following table:
 x 1 2 3 y 6 4 2

Plotting the points (1, 6), (2, 4) and (32) and drawing a line passing through these points, we obtain the graph of of the line  2x + y = 8.

The shaded region represents the area bounded by the lines x – y = 1, 2x + y = 8 and the y-axis. This represents a triangle.

It can be seen that the lines intersect at the point C(3, 2). Draw CD perpendicular from C on the y-axis.

Height = CD = 3 units

Base = AB = 9 units

∴ Area of the shaded region = Area of ∆ABC = $\frac{1}{2}×\mathrm{AB}×\mathrm{CD}=\frac{1}{2}×9×3$ = $\frac{27}{2}$ square units

#### Question 11:

Draw the graph for each of the equation x + y = 6 and x – y = 2 on the same graph paper and find the coordinates of the point where the two straight lines intersect.

$x+y=6\phantom{\rule{0ex}{0ex}}⇒y=-x+6$
When x = 0, $y=-0+6=6$
When x = 1, $y=-1+6=5$
When x = 3, $y=-3+6=3$

Thus, the points on the line x + y = 6 are as given in the following table:

 x 0 1 3 y 6 5 3

Plotting the points (0, 6), (1, 5) and (33) and drawing a line passing through these points, we obtain the graph of of the line x + y = 6.

$x-y=2\phantom{\rule{0ex}{0ex}}⇒y=x-2$
When x = 0, $y=0-2=-2$
When x = 2, $y=2-2=0$
When x = –1, $y=-1-2=-3$

Thus, the points on the line x – y = 2 are as given in the following table:
 x 0 2 –1 y –2 0 –3

Plotting the points (0, –2), (2, 0) and (–1–3) and drawing a line passing through these points, we obtain the graph of of the line x – y = 2.

It can be seen that the lines x + y = 6 and x – y = 2 intersect at the point (4, 2).

#### Question 12:

Two students A and B contributed ₹ 100 towards the Prime Minister's Relief Fund to help the earthquake victims. Write a linear equation to satisfy the above data and draw its graph.

Let the contribution of A and B be ₹ and ₹ y, respectively.

Total contribution of A and B = ₹ + ₹ = ₹ (y)

It is given that the total contribution of A and B is ₹ 100.

∴ y = 100

This is the linear equation satisfying the the given data.

= 100

⇒ y = 100 – x

When x = 10, y = 100 – 10 = 90

When x = 40, y = 100 – 40 = 60

When x = 60, y = 100 – 60 = 40

Thus, the points on the line = 100  are as given in the following table:

 x 10 40 60 y 90 60 40

Plotting the points (10, 90), (40, 60) and (6040) and drawing a line passing through these points, we obtain the graph of of the line = 100.

#### Question 1:

The equation of the x-axis is
(a)  x = 0
(b) y = 0
(c) x = y
(d) x + y = 0

Since, the y-coordinate of any point on x-axis is always 0.

So, the equation of the x-axis is y = 0.

Hence, the correct option is (b).

#### Question 2:

The equation of the y-axis is
(a)  x = 0
(b) y = 0
(c) x = y
(d) x + y = 0

Since, the x-coordinate of any point on y-axis is always 0.

So, the equation of the y-axis is x = 0.

Hence, the correct option is (a).

#### Question 3:

The point of the form (a, a), where a ≠ 0, lies on
(a) the x-axis
(b) the y-axis
(c) the line y = x
(d) the line x + y = 0

(c) the line y = x
Given, a point of the form , where $a$≠ 0 .
When , the point is (1,1)
When , the point is (2,2).....and so on.
Plot the points (1,1) and (2,2).......and so on. Join the points and extend them in both the direction. You will get the equation of the line .

#### Question 4:

The point of the form (a, a), where a ≠ 0, lies on
(a) the x-axis
(b) the y-axis
(c) the line y = x
(d) the line x + y = 0

(d) the line x + y = 0
Given, a point of the form , where $a$ ≠ 0.
When , the point is (1,-1).
When , the point is (2,-2).
When , the point is (3,-3).......and so on.
Plot these points on a graph paper. Join these points and extend them in both the directions.
You will get the equation of the line .

#### Question 5:

The linear equation 3x − 5y = has
(a) a unique solution
(b) two solutions
(c) infinitely many solutions
(d) no solution

(c) infinitely many solutions
Given linear equation:
Or,

When , .

When  , .

When , .

Thus, we have the following table:

 $\text{x}$ 5 10 0 $y$ 0 3 -3

Plot the points  and $C\left(0,-3\right)$. Join the points and extend them in both the directions.
We get infinite points that satisfy the given equation.
Hence, the linear equation has infinitely many solutions.

#### Question 6:

The equation 2x + 5y = 7 has a unique solution, if x and y are
(a) natural numbers
(b) rational numbers
(c) positive real numbers
(d) real numbers

Since, every point on the line represented by the equation 2x + 5= 7 is its solution.

Therefore, there are infinite solutions of the equation the equation 2x + 5= 7 in which the values of x and y are rational numbers, positive real numbers or real numbers.

But, as 2 + 5 = 7, i.e. x = 1 and y = 1 are the only pair of natural numbers that are the solution of the equation the equation 2x + 5= 7.

So, the equation 2x + 5= 7 has a unique solution, if x and are both are natural numbers.

Hence, the correct option is (a).

#### Question 7:

The graph of y = 5 is a line
(a) making an intercept 5 on the x-axis
(b) making an intercept 5 on the y-axis
(c) parallel to the x-axis at a distance of 5 units from the origin
(d) parallel to the y-axis at a distance of 5 units from the origin

As, the graph of y = 5 is a line parallel to x-axis i.e. y = 0.

$⇒$ The line represented by the equation y = 5 is parallel to x-axis and intersects y-axis at y = 5.

So, the graph of y = 5 is a line parallel to the x-axis at a distance of 5 units from the origin making an intercept 5 on the y-axis.

Hence, the correct answers are options (b) and (c).
Disclaimer: In this question, there are two correct answers.

#### Question 8:

The graph of x = 4 is a line
(a) making an intercept 4 on the x-axis
(b) making an intercept 4 on the y-axis
(c) parallel to the x-axis at a distance of 4 units from the origin
(d) parallel to the y-axis at a distance of 4 units from the origin

As, the graph of x = 4 is a line parallel to y-axis i.e. x = 0.

$⇒$ The line represented by the equation x = 4 is parallel to y-axis and intersects x-axis at x = 4.

So, the graph of x = 4 is parallel to y-axis at a distance of 4 units from the origin making an intercept 4 on the x-axis.

Hence, the correct options are (a) and (d).

Disclaimer: In this question, there are two correct answers.

#### Question 9:

The graph of x + 3 = 0 is a line
(a) making an intercept –3 on the x-axis
(b) making an intercept –3 on the y-axis
(c) parallel to the y-axis at a distance of 3 units to the left of y-axis
(d) parallel to the x-axis at a distance of 3 units below the x-axis

As, the graph of x + 3 = 0 or x = $-$3 is a line parallel to y-axis i.e. x = 0.

$⇒$ The line represented by the equation x = $-$3 is parallel to y-axis and intersects x-axis at x = $-$3.

So, the graph of x + 3 = 0 is a line parallel to the y-axis at a distance of 3 units to the left of y-axis making an intercept $-$3 on the x-axis.

Hence, the correct options are (a) and (c).

Disclaimer: In this question, there are two correct answers.

#### Question 10:

The graph of y + 2 = 0 is a line
(a) making an intercept –2 on the x-axis
(b) making an intercept –2 on the y-axis
(c) parallel to the x-axis at a distance of 2 units below the x-axis
(d) parallel to the y-axis at a distance of 2 units to the left of y-axis

As, the graph of y + 2 = 0 or y = −2 is a line parallel to x-axis i.e. y = 0.

⇒ The line represented by the equation y = −2 is parallel to x-axis and intersects y-axis at y = −2.

So, the graph of y + 2 = 0 is a line parallel to the x-axis at a distance of 2 units below the x-axis making an intercept −2 on the y-axis.

Hence, the correct options are (b) and (c).

Disclaimer: In this question, there are two correct answers.

#### Question 11:

The graph of the linear equation 2x + 3y = 6 meets the y-axis at the point
(a) (2, 0)
(b) (3, 0)
(c) (0, 2)
(d) (0, 3)

If he graph of the linear equation 2x + 3= 6 meets the y-axis, then x = 0.

Substituting the value of x = 0 in equation 2x + 3= 6, we get

$2\left(0\right)+3y=6\phantom{\rule{0ex}{0ex}}⇒3y=6\phantom{\rule{0ex}{0ex}}⇒y=\frac{6}{3}\phantom{\rule{0ex}{0ex}}⇒y=2$

So, the point of meeting is (0, 2).

Hence, the correct answer is option (c).

#### Question 12:

The graph of the linear equation 2x + 5y = 10 meets the x-axis at the point
(a) (0, 2)
(b) (2, 0)
(c) (5, 0)
(d) (0, 5)

If he graph of the linear equation 2x + 5= 10 meets the x-axis, then y = 0.

Substituting the value of y = 0 in equation 2x + 5= 10, we get

$2x+5\left(0\right)=10\phantom{\rule{0ex}{0ex}}⇒2x=10\phantom{\rule{0ex}{0ex}}⇒x=\frac{10}{2}\phantom{\rule{0ex}{0ex}}⇒x=5$

So, the point of meeting is (5, 0).

Hence, the correct option is (c).

#### Question 13:

The graph of the line x = 3 passes through the point
(a) (0, 3)
(b) (2, 3)
(c) (3, 2)
(d) None of these

(c) (3, 2)
The graph of line x = 3 is a line parallel to the y-axis.
Hence, its passes through (3,2), satisfying x =3.

#### Question 14:

The graph of the line y = 3 passes through the point
(a) (3, 0)
(b) (3, 2)
(c) (2, 3)
(d) none of  these

Since, the graph of the line y = 3 is parallel to x-axis at a distance of 3 units from the x-axis.

Or, the y-coordinate of every point on the line is always equal to 3.

So, the graph of the line y = 3 passes through the point (2, 3).

Hence, the correct option is (c).

#### Question 15:

The graph of the line y = −3 does not pass through the point
(a) (2, −3)
(b) (3, −3)
(c) (0, −3)
(d) (−3, 2)

(d) (−3, 2)

The graph of the line y = -3 does not pass through (-3,2) since (-3,2) does not satisfy y = -3.

#### Question 16:

The graph of the linear equation x y = 0 passes through the point
(a)
(b)
(c) (0, −1)
(d) (1, 1)

(d) (1, 1)
Given equation: or,
When ,
When , ... and so on
Thus, we get the following table:

 $\text{x}$ 1 2 $y$ 1 2

​Plot the points on the graph paper. Join the points and extend them in both the directions.
We can see the linear equation  passes through the point (1,1).

#### Question 17:

Each of the points (−2, 2), (0, 0), (2, −2) satisfies the linear equation
(a) xy = 0
(b) x + y = 0
(c) −x + 2y = 0
(d) x – 2y = 0

(b) x + y = 0
Given points:  and
We have to check which equation satisfies the given points.
Let us check for (a)
Substituting in the equation, we get:
Substituting  in the equation, we get:
Substituting  in the equation, we get:
So, the given points do not satisfy the equation.
Now, let us check (b)
Substituting  in the equation, we get:
Substituting  in the equation, we get:
Substituting $\left(2,-2\right)$ in the equation, we get:
So, the given points satisfy the equation.
Similarly, we can check other equations and find that the given points will not satisfy the equations.
Hence, each of  and  is a solution of the linear equation .

#### Question 18:

How many linear equations in x and y can be satisfied by x = 2, y = 3?
(a) Only one
(b) Only two
(c) Infinitely many
(d) None of these

(c) Infinitely many
Infinite linear equations are satisfied by .

#### Question 19:

A linear equation in two variables x and y 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

(a) a ≠ 0, b ≠ 0
A linear equation in two variables $\text{x}$ and $y$ is of the form , where $a$ ≠ 0 and $b$ ≠ 0.

#### Question 20:

If (2, 0) is a solution of the linear equation 2x + 3y = k then the value of k is
(a) 6
(b) 5
(c) 2
(d) 4

Since, (2, 0) is a solution of the linear equation 2x + 3y = k.

Substituting the values x = 2 and y = 0 in the equation 2x + 3y = k, we get

Hence, the correct option is (d).

#### Question 21:

Any point on the x-axis is of the form
(a) (x, y)
(b) (0, y)
(c) (x, 0)
(d) (x, x)

(c) (x, 0),
Any point on the $\text{x}$-axis is of the form $\left(\text{x,0)}$, where $\text{x}$ ≠ 0.

#### Question 22:

Any point on the y-axis is of the form
(a) (x, y)
(b) (0, y)
(c) (x, 0)
(d) (y, y)

(b) (0, y),
Any point on the $y$-axis is of the form $\left(0,y\right)$, where $y$ ≠ 0.

#### Question 23:

x = 5, y = 2 is a solution of the linear equation
(a) x + 2y = 7
(b) 5x + 2y = 7
(c) x + y = 7
(d) 5x + y = 7

Substituting the values x = 5, y = 2 in

(a) x + 2y = 7, we get

(b) 5x + 2y = 7, we get

(c) x + y = 7, we get

(d) 5x + y = 7, we get

Hence, the correct option is (c).

#### Question 24:

If the point (3, 4) lies on the graph of 3y = ax + 7 then the value of a is

(a) $\frac{2}{5}$

(b) $\frac{5}{3}$

(c) $\frac{3}{5}$

(d) $\frac{2}{7}$