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#### Question 1:

Draw the graph of each of the following linear equations in two variables:

(i) x + y = 4

(ii) x − y = 0

(iii) −x + y = 6

(iv) y = 2x

(v) 3x + 5y = 15

(vi) $\frac{x}{2}-\frac{y}{3}=2$

(vii)

(viii) 2y = −x + 1

(i) We are given,

x + y = 4

We get,

y = 4 – x,

Now, substituting x = 0 in y = 4 – x, we get

y = 4

Substituting x = 4 in y = 4 – x, we get

y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given

 x 0 4 y 4 0

(ii) We are given,

We get,

Now, substituting x = 0 in y = x – 2, we get

Substituting x = 2 in y = x – 2, we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 2 y –2 0

(iii)
We are given,

We get,

Now, substituting in ,we get

Substituting x = –6 in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 −6 y 6 0

(iv) We are given,

Now, substituting x = 1 in ,we get

Substituting x = 3 in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 1 3 y 2 6

(v) We are given,

We get,

Now, substituting x = 0 in ,we get

Substituting x = 5 in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 5 y 3 0

(vi) We are given,

We get,

Now, substituting x = 0 in ,we get

Substituting x = 4 in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 4 y –6 0

(vii) We are given,

We get,

Now, substituting x = 5 in ,we get

Substituting x = 8 in ,we get

y = 5

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 5 8 y 4 5

(viii) We are given,

We get,

Now, substituting x = 1 in ,we get

Substituting x = 5 in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 1 5 y –1 –2

#### Question 2:

Give the equations of two lines passing through (3, 12). How many more such lines are there, and why?

We observe that x = 3 and y = 12 is the solution of the following equations

So, we get the equations of two lines passing through (3, 12) are, 4xy = 0 and 3xy + 3 = 0.

We know that passing through the given point infinitely many lines can be drawn. So, there are infinitely many lines passing through (3,12)

#### Question 3:

A three-wheeler scooter charges Rs 15 for first kilometer and Rs 8 each for every subsequent kilometer. For a distance of x km, an amount of Rs y is paid. Write the linear equation representing the above information.

Total fare of Rs y for covering the distance of x km is given by
y = 15 + 8(x − 1)
y = 15 + 8x − 8
y = 8x + 7

Where, Rs y is the total fare (x – 1) is taken as the cost of first kilometer is already given Rs 15 and 1 has to subtracted from the total distance travelled to deduct the cost of first kilometer.

#### Question 4:

Plot the points (3,5) and (−1, 3) on a graph paper and verify that the straight line passing through these points also passes through the point (1, 4).

The required graph is below:-

By plotting the given points (3, 5) and (–1, 3) on a graph paper, we get the line BC.

We have already plotted the point A (1, 4) on the given plane by the intersecting lines.

Therefore, it is proved that the straight line passing through (3, 5) and (–1, 3) also passes through A (1, 4).

#### Question 5:

From the choices given below, choose the equation whose graph is given in the figure.

(i) y = x

(ii) x + y = 0

(iii) y = 2x

(iv) 2 + 3y = 7x

We are given co-ordinates (1, –1) and (–1, 1) as the solution of one of the following equations.

We will substitute the value of both co-ordinates in each of the equation and find the equation which satisfies the given co-ordinates.

(i) We are given,

Substituting ,we get

Substituting ,we get

Therefore, the given equationdoes not represent the graph in the figure.

(ii) We are given,

Substituting ,we get

Substituting ,we get

Therefore, the given solutions satisfy this equation. Thus, it is the equation whose graph is given.

#### Question 6:

From the choices given below, choose the equation whose graph is given in the figure.

(i) y= x + 2

(ii) y = x −2

(iii) y = x + 2

(iv) x + 2y = 6

We are given co-ordinates (–1, 3) and (2, 0) as the solution of one of the following equations.

We will substitute the value of both co-ordinates in each of the equation and find the equation which satisfies the given co-ordinates.

(i) We are given,

Substituting ,we get

Substituting ,we get

Therefore, the given solutions does not satisfy this equation.

(ii) We are given,

Substituting ,we get

Substituting ,we get

Therefore, the given solutions does not completely satisfy this equation.

(iii) We are given,

Substituting ,we get

Substituting ,we get

Therefore, the given solutions satisfy this equation. Thus, it is the equation whose graph is given.

#### Question 7:

If the point (2, −2) lies on the graph of the linear equation 5x + ky = 4,  find the value of k.

It is given that the point lies on the given equation,

Clearly, the given point is the solution of the given equation.

Now,

Substituting in the given equation, we get

#### Question 8:

Draw the graph of the equation 2x + 3y = 12. From the graph, find the coordinates of the point:

(i) whose y-coordinates is 3.

(ii) whose x-coordinate is −3.

We are given,

We get,

Substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 6 y 4 0

By plotting the given equation on the graph, we get the point B (0, 4) and C (6,0).

(i) Co-ordinates of the point whose y axis is 3 are A

(ii) Co-ordinates of the point whose x -coordinate is –3 are D

#### Question 9:

Draw the graph of each of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes:

(i) 6x − 3y = 12

(ii) −x + 4y = 8

(iii) 2x + y = 6

(iv) 3x + 2y + 6 = 0

(i) We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 2 y –4 0

Co-ordinates of the points where graph cuts the co-ordinate axes are at y axis and

at x axis.

(ii) We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 –8 y 2 0

Co-ordinates of the points where graph cuts the co-ordinate axes are at y axis and

at x axis.

(iii) We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 3 y 6 0

Co-ordinates of the points where graph cuts the co-ordinate axes are at y axis and

at x axis.

(iv) We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 –2 y –3 0

Co-ordinates of the points where graph cuts the co-ordinate axes are at y axis and

at x axis.

#### Question 10:

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Aarushi paid Rs 27 for a book kept for seven days. If fixed charges are Rs x and per day charges are Rs y. Write the linear equation representing the above information.

Total charges of Rs 27 of which Rs x for first three days and Rs y per day for 4 more days is given by

Here, is taken as the charges for the first three days are already given at Rs x and we have to find the charges for the remaining four days as the book is kept for the total of 7 days.

#### Question 11:

A number is 27 more than the number obtained by reversing its digits. If its unit's and ten's digit are x and y respectively, write the linear equation representing the above statement.

The number given to us is in the form of yx,

where y represents the ten’s place of the number

And x represents the unit’s place of the number

Now, the given number is

number obtained by reversing the digits of the number is

It is given to us that the original number is 27 more than the number obtained by reversing its digits

So,

#### Question 12:

The sum of a two digit number and the number obtained by reversing the order of its digits is 121. If units and ten's digit of the number are x and y respectively, then write the linear equation representing the above statement.

The number given to us is in the form of yx.

where y represents the tens place of the number

And x represents the units place of the number

Now, the given number is

number obtained by reversing the digits of the number is

It is given to us that the sum of these two numbers is 121

So,

#### Question 13:

Draw the graph of the equation 2x + y = 6. Shade the region bounded by the graph and the coordinate axes. Also, find the area of the shaded region.

We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 3 y 6 0

The region bounded by the graph is ABC which forms a triangle.

AC at y axis is the base of triangle having AC = 6 units on y axis.

BC at x axis is the height of triangle having BC = 3 units on x axis.

Therefore,

Area of triangle ABC, say A is given by

#### Question 14:

Draw the graph of the equation $\frac{x}{3}+\frac{y}{4}=1$. Also, find the area of the triangle formed by the line and the coordinates axes.

We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 3 y 4 0

The region bounded by the graph is ABC which form a traingle.

AC at y axis is the base of traingle having AC = 4 units on y axis.

BC at x axis is the height of traingle having BC = 3 units on x axis.

Therefore,

Area of traingle ABC,say A is given by

#### Question 15:

Draw the graph of y = | x |.

We are given,

Substituting, we get

Substituting, we get

Substituting, we get

Substituting, we get

For every value of x, whether positive or negative, we get y as a positive number.

#### Question 16:

Draw the graph of y = | x | + 2.

We are given,

Substituting, we get

Substituting, we get

Substituting, we get

Substituting, we get

Substituting, we get

For every value of x, whether positive or negative, we get y as a positive number and the minimum value of y is equal to 2 units.

#### Question 17:

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 area bounded by these lines and x-axis.

We are given,

We get,

Now, substituting in , we get

Substituting in , we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 6 y 4 0

Plotting A(0,4) and E(6,0) on the graph and by joining the points , we obtain the graph of equation .

We are given,

We get,

$y=x-1$

Now, substituting in $y=x-1$,we get

$y=-1$

Substituting in $y=x-1$,we get

$y=-2$

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 –1 y -1 -2

Plotting D(0,1) and E(-1,0) on the graph and by joining the points , we obtain the graph of equation .

By the intersection of lines formed by and on the graph, triangle ABC is formed on y axis.

Therefore,

AC at y axis is the base of triangle ABC having AC = 5 units on y axis.

Draw FE perpendicular from F on y axis.

FE parallel to x axis is the height of triangle ABC having FE = 3 units on x axis.

Therefore,

Area of triangle ABC, say A is given by

#### Question 18:

Draw the graphs of the linear equations 4x − 3y + 4 = 0 and 4x + 3y −20 = 0. Find the area bounded by these lines and x-axis.

We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 –1 y 0

Plotting E(0, ) and A(-1,0) on the graph and by joining the points , we obtain the graph of equation .

We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 5 y 0

Plotting D(0, ) and B(5,0) on the graph and by joining the points , we obtain the graph of equation .

By the intersection of lines formed by and on the graph, triangle ABC is formed on x axis.

Therefore,

AB at x axis is the base of triangle ABC having AB = 6 units on x axis.

Draw CF perpendicular from C on x axis.

CF parallel to y axis is the height of triangle ABC having CF = 4 units on y axis.

Therefore,

Area of triangle ABC, say A is given by

#### Question 19:

The path of a train A is given by the equation 3x + 4y − 12 = 0 and the path of another train B is given by the equation 6x + 8y − 48 = 0. Represent this situation graphically.

We are given the path of train A,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 4 y 3 0

Plotting A(4,0) and E(0,3) on the graph and by joining the points , we obtain the graph of equation .

We are given the path of train B,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 8 y 6 0

Plotting C(0,6) and D(8,0) on the graph and by joining the points , we obtain the graph of equation

#### Question 20:

Ravish tells his daughter Aarushi, ''Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be''. If present ages of Aarushi and Ravish are x and y years respectively, represent this situation algebraically as well as graphically.

We are given the present age of Ravish as y years and Aarushi as x years.

Age of Ravish seven years ago

Age of Aarushi seven years ago

It has already been said by Ravish that seven years ago he was seven times old then Aarushi was then

So,

Age of Ravish three years from now

Age of Aarushi three years from now

It has already been said by Ravish that three years from now he will be three times old then Aarushi will be then

So,

(1) and (2) are the algebraic representation of the given statement.

We are given,

We get,

Now, substituting in ,we get

Substituting in, we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 6 y –42 0

We are given,

We get,

Now, substituting in ,we get

Substituting in ,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 –2 y 6 0

The red -line represents the equation.

The blue-line represents the equation.

#### Question 21:

Aarushi was driving a car with uniform speed of 60km/h. Draw distance-time graph. From the graph, find the distance travelled by Aarushi in

(i) $2\frac{1}{2}$ Hours

(ii) $\frac{1}{2}$ Hour

Aarushi is driving the car with the uniform speed of 60 km/h.

We represent time on X-axis and distance on Y-axis

Now, graphically

We are given that the car is travelling with a uniform speed 60 km/hr. This means car travels 60 km distance each hour. Thus the graph we get is of a straight line.

Also, we know when the car is at rest, the distance travelled is 0 km, speed is 0 km/hr and the time is also 0 hr.
Thus, the given straight line will pass through O(0,0) and M(1,60).

Join the points O and M and extend the line in both directions.

Now, we draw a dotted line parallel to y-axis from x = $\frac{1}{2}$ that meets the straight line graph at L from which we draw a line parallel to x-axis that crosses y-axis at 30. Thus, in $\frac{1}{2}$hr, distance travelled by the car is 30 km.

Now, we draw a dotted line parallel to y-axis from x = $2\frac{1}{2}$ that meets the straight line graph at N from which we draw a line parallel to x-axis that crosses y-axis at 150. Thus, in $2\frac{1}{2}$hr, distance travelled by the car is 150 km.

(i)

Distance travelled in hours is given by

(ii)

Distance travelled in hours is given by

#### Question 1:

Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:

(i) −2x + 3y = 12

(ii) $x-\frac{y}{2}-5=0$

(iii) 2x + 3y = 9.35

(iv) 3x = −7y

(v) 2x + 3 = 0

(vi) y − 5 = 0

(vii) 4 = 3x

(viii) $y=\frac{x}{2}$

(i) We are given

Comparing the given equation with ,we get

(ii) We are given

Comparing the given equation with ,we get

(iii) We are given

Comparing the given equation with ,we get

(iv) We are given

Comparing the given equation with ,we get

(v) We are given

Comparing the given equation with ,we get

(vi) We are given

Comparing the given equation with ,we get

(vii) We are given

Comparing the given equation with ,we get

(viii) We are given

Comparing the given equation with ,we get

#### Question 2:

Write each of the following as an equation in two variables:

(i) 2x =−3

(ii) y = 3

(iii) 5x = $\frac{7}{2}$

(iv) y = $\frac{3}{2}x$

(i) We are given,

Now, in two variable forms the given equation will be

(ii) We are given,

Now, in two variable forms the given equation will be

(iii) We are given,

Now, in two variable forms the given equation will be

(iv) We are given,

Now, in two variable forms the given equation will be

#### Question 3:

The cost of ball pen is Rs 5 less than half of the cost of fountain pen. Write this statement as a linear equation in two variables.

Let the cost of fountain pen be x and cost of ball pen be y.

According to the given equation, we have

Here x is the cost of one fountain pen and y is that of one ball pen.

#### Question 1:

Give the geometric representations of the following equations

(a) on the number line
(b) on the Cartesian plane:

(i) x = 2

(ii) y + 3 = 0

(iii) y = 3

(iv) 2x + 9 = 0

(v) 3x − 5 = 0

(i) We are given,

x = 2

The representation of the solution on the number line, when given equation is treated as an equation in one variable.

The representation of the solution on the Cartesian plane, it is a line parallel to y axis passing through the point (2, 0) is shown below

(ii) We are given,

We get,

The representation of the solution on the number line, when given equation is treated as an equation in one variable.

The representation of the solution on the Cartesian plane, it is a line parallel to x axis passing through the point A(0, –3) is shown below

(iii) We are given,

The representation of the solution on the number line, when given equation is treated as an equation in one variable.

The representation of the solution on the Cartesian plane, it is a line parallel to x axis passing through the point (0, 3) is shown below

(iv) We are given,

We get,

The representation of the solution on the number line, when given equation is treated as an equation in one variable.

The representation of the solution on the Cartesian plane,it is a line parallel to y axis passing through the point is shown below

(v) We are given,

We get,

The representation of the solution on the number line, when given equation is treated as an equation in one variable.

The representation of the solution on the Cartesian plane, it is a line parallel to y axis passing through the point is shown below

#### Question 2:

Give the geometrical representation of 2x + 13 = 0 as an equation in

(i) on variable

(ii) two variable

We are given,

We get,

The representation of the solution on the number line, when given equation is treated as an equation in one variable.

The representation of the solution on the Cartesian plane, it is a line parallel to y axis passing through the point is shown below

#### Question 3:

Solve the equation 3x + 2 = x − 8, and represent the solution on (i) the number line
(ii) the Cartesian plane.

We are given,

we get,

The representation of the solution on the number line, when given equation is treated as an equation in one variable.

The representation of the solution on the Cartesian plane, it is a line parallel to y axis passing through the point (–5, 0) is shown below

#### Question 4:

Write the equation of the line that is parallel to x-axis and passing through the point.

(i) (0,3)

(ii) (0, −4)

(iii) (2, −5)

(iv) (3, 4)

(i) We are given the co-ordinates of the Cartesian plane at (0,3).

For the equation of the line parallel to x axis ,we assume the equation as a one variable equation independent of x containing y equal to 3.

We get the equation as

(ii) We are given the co-ordinates of the Cartesian plane at (0,-4).

For the equation of the line parallel to x axis ,we assume the equation as a one variable equation independent of x containing y equal to -4.

We get the equation as

(iii) We are given the co-ordinates of the Cartesian plane at (2,-5).

For the equation of the line parallel to x axis ,we assume the equation as a one variable equation independent of x containing y equal to -5.

We get the equation as

(iv) We are given the co-ordinates of the Cartesian plane at (3,4).

For the equation of the line parallel to x axis ,we assume the equation as a one variable equation independent of x containing y equal to 4.

We get the equation as

#### Question 5:

Write the equation of the line that is parallel to y-axis and passing through the point

(i) (4,0)

(ii) (−2,0)

(iii) (3, 5)

(iv) (−4, −3)

(i) We are given the co-ordinates of the Cartesian plane at (4,0).

For the equation of the line parallel to y axis ,we assume the equation as a one variable equation independent of y containing x equal to 4.

We get the equation as

(ii) We are given the co-ordinates of the Cartesian plane at (–2,0).

For the equation of the line parallel to y axis ,we assume the equation as a one variable equation independent of y containing x equal to –2.

We get the equation as

(iii) We are given the co-ordinates of the Cartesian plane at (3,5).

For the equation of the line parallel to y axis, we assume the equation as a one variable equation independent of y containing x equal to 3.

We get the equation as

(iv) We are given the co-ordinates of the Cartesian plane at (−4,−3).

For the equation of the line parallel to y axis, we assume the equation as a one variable equation independent of y containing x equal to −4.

We get the equation as

#### Question 1:

Mark the correct alternative in each of the following:

If (4, 19) is a solution of the equation y = ax + 3, then a=

(a) 3

(b) 4

(c) 5

(d) 6

We are given (4, 19)as the solution of equation

Substituting x = 4 and y = 19, we get

Therefore, the correct answer is (b).

#### Question 2:

If (a, 4) lies on the graph of 3x + y = 10, then the value of a is

(a) 3

(b) 1

(c) 3

(d) 4

We are given (a, 4) lies on the graph of linear equation 3x + y = 10.

So, the given co-ordinates are the solution of the equation 3x + y = 10.

Therefore, we can calculate the value of a by substituting the value of given co-ordinates in equation 3x + y = 10.

Substituting x = a and y = 4 in equation 3x + y = 10, we get

No option is correct.

#### Question 3:

The graph of the linear equation 2xy = 4 cuts x- axis at

(a) (2, 0)

(b) (−2, 0)

(c) (0, −4)

(d) (0, 4)

We are given,

we get,

We will substitute in to get the co-ordinates for the graph of at x axis

Co-ordinates for the graph of are .

Therefore, the correct answer is (a).

#### Question 4:

How many linear equations are satisfied by x = 2  and y = −3?

(a) Only one

(b) Two

(c) Three

(d) Infinitely many

There are infinite numbers of linear equations that are satisfied by as

(i) Every solution of the linear equation represent a point on the line.

(ii) Every point on the line is the solution of the linear equation.

Therefore, the correct answer is (d).

#### Question 5:

The equation x − 2 = 0 on number line is represented by

(a) a line

(b) a point

(c) infinitely many lines

(d) two lines

The equation is represented by a point on the number line.

Therefore, the correct answer is (b).

#### Question 6:

x = 2, y = −1 is a solution of the linear equation

(a) x + 2y = 0

(b) x + 2y = 4

(c) 2x + y = 0

(d) 2x + y = 5

We are given as the solution of linear equation, which we have to find?

The equation is which can be proved by

Substituting in the equation, we get

Therefore, the correct answer is (a).

#### Question 7:

If (2k − 1, k) is a solution of the equation 10x − 9y = 12, then k =

(a) 1

(b) 2

(c) 3

(d) 4

We are given as the solution of equation

Substituting, we get

Therefore, the correct answer is (b).

#### Question 8:

The distance between the graph of the equations x = −3 and x = 2 is

(a) 1

(b) 2

(c) 3

(d) 5

Distance between the graph of equations, say D

D = Distance of co-ordinate on negative side of x axis + Distance of co-ordinate on positive side of x axis

Distance of co-ordinate on negative side of x axis = x = 3 units

Distance of co-ordinate on positive side of x axis = x = 2 units

Therefore, the correct answer is (d).

#### Question 9:

The distance between the graphs of the equations y = −1 and y = 3 is

(a) 2

(b) 4

(c) 3

(d) 1

Distance between the graph of equations, say D

D = Distance of co-ordinate on negative side of y axis + Distance of co-ordinate on positive side of y axis

Distance of co-ordinate on negative side of y axis = y = 1 units

Distance of co-ordinate on positive side of y axis = y = 3 units

Therefore, the correct answer is (b).

#### Question 10:

If the graph of the equation 4x + 3y = 12 cuts the coordinate axes at A and B, then hypotenuse of right triangle AOB is of length

(a) 4 units

(b) 3 units

(c) 5 units

(d) none of these

(i) We are given,

We get,

Now, substituting in, we get

Substituting in,we get

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

 x 0 3 y 4 0

We are given that the graph of equation cuts the co-ordinate axes and and forms the right angle triangle AOB.

Length of AB in right angled triangle AOB

The length of hypotenuse AB of triangle AOB is 5 units.

Therefore, the correct answer is (c).

#### Question 1:

Write the equation representing x-axis.

The equation of line representing x axis is given by

#### Question 2:

Write the equation representing y-axis.

The equation of line representing y axis is given by

#### Question 3:

Write the equation of a line passing through the point (0, 4) and parallel to x-axis.

We are given the co-ordinates of the Cartesian plane at (0,4).

For the equation of the line parallel to x axis, we assume the equation as a one variable equation independent of x containing y equal to 4.

We get the equation as

#### Question 4:

Write the equation of a line passing through the point (3, 5)  and parallel to x-axis.

We are given the co-ordinates of the Cartesian plane at (3,5).

For the equation of the line parallel to x axis, we assume the equation as a one variable equation independent of x containing y equal to 5.

We get the equation as

#### Question 5:

Write the equation of a line parallel to y-axis and passing through the point (−3, −7).

We are given the co-ordinates of the Cartesian plane at (–3,–7).

For the equation of the line parallel to y axis, we assume the equation as a one variable equation independent of y containing x equal to –3.

We get the equation as

#### Question 6:

A line passes through the point (−4, 6) and is parallel to x-axis. Find its equation.

We are given the co-ordinates of the Cartesian plane at (–4,6).

For the equation of the line parallel to x axis, we assume the equation as a one variable equation independent of x containing y equal to 6.

We get the equation as

#### Question 7:

Solve the equation 3x − 2 = 2x + 3 and represent the solution on the number line.

We are given,

we get,

The representation of the solution on the number line, when given equation is treated as an equation in one variable.

#### Question 8:

Solve the equation 2y − 1 = y + 1 and represent it graphically on the coordinate plane.

We are given,

we get,

The representation of the solution on the Cartesian plane, it is a line parallel to y axis passing through the point is shown below

#### Question 9:

If the point (a, 2) lies on the graph of the linear equation 2x − 3y + 8 = 0, find the value of a.

We are given lies on the graph of linear equation.

So, the given co-ordinates are the solution of the equation.

Therefore, we can calculate the value of a by substituting the value of given co-ordinates in equation.

Substituting in equation, we get

#### Question 10:

Find the value of k for which the point (1, −2) lies on the graph of the linear equation x − 2y + k = 0.

We are given lies on the graph of linear equation.

So, the given co-ordinates are the solution of the equation.

Therefore, we can calculate the value of k by substituting the value of given co-ordinates in equation.

Substituting in equation, we get

#### Question 1:

Write two solutions for each of the following equations:

(i) 3x + 4y = 7

(ii) x = 6y

(iii) x + xy = 4

(iv) $\frac{2}{3}x-y=4$

(i) We are given,

Substituting x = 1 in the given equation, we get

Substituting x = 2 in the given equation, we get

(ii) We are given,

Substituting in the given equation, we get

Substituting in the given equation, we get

(iii) We are given,

Substituting x = 0 in the given equation, we get

Substituting x = 4 in the given equation, we get

(iv) We are given,

Substituting x = 0 in the given equation, we get

Substituting x = 3 in the given equation, we get

#### Question 2:

Check which of the following are solutions of the equations 2x − y = 6 and which are not:

(i) (3,0)

(ii) (0,6)

(iii) (2, −2)

(iv)

(v)

We are given,

2xy = 6

(i) In the equation 2xy = 6,we have

Substituting x = 3 and y = 0 in 2xy, we get

is the solution of 2x – y = 6.

(ii) In the equation 2x – y = 6, we have

Substituting x = 0 and y = 6 in 2x – y,we get

is not the solution of 2x – y = 6.

(iii) In the equation 2x – y = 6,we have

Substituting x = 2 and y = –2  in 2xy, we get

is the solution of 2x – y = 6.

(iv) In the equation 2x – y = 6, we have

Substituting and y = 0 in 2x – y, we get

is not the solution of 2x – y = 6.

(v) In the equation 2x – y = 6, we have

Substituting and y = –5 in 2x – y, we get

is the solution of 2x – y = 6.

#### Question 3:

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

We are given,

is the solution of equation .

Substituting and in ,we get

#### Question 4:

Find the value of λ, if x  = −λ and y = $\frac{5}{2}$ is a solution of the equation x + 4y − 7 = 0.

We are given,

is the solution of equation .

Substituting and in ,we get

#### Question 5:

If x = 2α + 1 and y = α − 1 is a solution of the equation 2x − 3y + 5 = 0, find the value of α.

We are given,

is the solution of equation .

Substituting and in ,we get

#### Question 6:

If x = 1 and y = 6 is a solution of the equation 8xay + a2 = 0, find the value of a.

We are given,

is the solution of equation .

Substituting and in ,we get

#### Question 7:

Write two solutions of the form x = 0, y = a and x = b, y = 0 for each of the following equations:

(i) 5x − 2y = 10

(ii) −4x + 3y = 12

(iii) 2x + 3y =24

(i) We are given,

Substituting x = 0 in the given equation, we get

Substituting y = 0 in the given equation, we get

(ii) We are given,

Substituting in the given equation, we get
$-4×0+3y=12\phantom{\rule{0ex}{0ex}}3y=12\phantom{\rule{0ex}{0ex}}y=4$
Thus x = 0 and y = 4 is the solution of the −4x + 3y = 12

Substituting y = 0 in the given equation, we get

(iii) We are given,

Substituting x = 0 in the given equation, we get

Substituting y = 0 in the given equation, we get

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