Sponsor Area
Let the cost of a notebook be Rs x and the cost of a pen be Rs y.
Then, according to the given condition of the question,
x = 2y
⇒ x – 2y = 0
This is the required linear equation in two variables x and y.
, c = - 10
- 2x + 3y = 6
- 2x + 3y - 6 = 0
Comparing with ax + by + c = 0, we get
a = - 2, b = 3, c = -6
x = 3y
x - 3y + 0 = 0
Comparing with ax + by + c = 0, we get
a = 1, b = -3, c = 0
2x = - 5y
2x + 5y + 0 = 0
Comparing with ax + by + c = 0, we get
a = 2, b = -5, c = 0
3x + 2 = 0
3x + 0y + 2 = 0
Comparing with ax + by + c = 0, we get
a = 3, b = 0, c = 2
y - 2 = 0 
0.x + 1.y - 2 = 0
Comparing with ax + by + c = 0, we get
a = 0, b = 1, c = -2
5 = 2x
-2x + 5 = 0
-2x + 0.y + 5 = 0
Comparing with ax + by + c = o, we get a = - 2, b = 0, c = 5
Let the present ages of father and son be x years and y years respectively.
Then,
Age of father 5 years ago = (x – 5) years
Age of his son 5 years ago = (y – 5) years
According to the question,
x – 5 = 7(y – 5) + 2
⇒ x – 5 = 7y – 35 + 2
⇒ x – 7y + 28 = 0
which is the required linear equation in two variables.
The given equation is
2x = y ⇒ 2x – y
⇒ 2x – y + 0 = 0
Comparing with ax + by + c = 0, we get
a = 2
b = –1
c = 0
This equation has infinitely many solutions.
= 2y – 3 and lx + my – n = 0 and write the value of l, m and n.
7 = 2x
⇒ 2x – 7 = 0
⇒ 2x + 0y – 7 = 0
Comparing with ax + by + c = 0, we get
a = 2
b = 0
c = –7
Solution not provided.
Ans. (i) 2X + 3Y - 4.37 = 0; a = 2, b = 3 and c = - 4.37
(ii) x - 
- 4 = 0; s = 1, b = 
c = -4
(iii) - 5x + 3y + 4 = 0; a = -5, b = 3 and c = 4
or
5x - 3y - 4 = 0; a = 5, b = 3 and c = -4
(iv) 2x - y + 0 = 0; a = 2, b = -1 and c = 0
Write each of the following as an equation in two variables:
(i) x = – 5 (ii) y = 2
(iii) 2x = 3 (iv) 5y = 2.
Solution not provided.
Ans. (i) 1 x + 0y + 5 = 0
(ii) 0x + 1y - 2 = 0
(iii) 2x + 0y - 3 = 0
(iv) 0x + 5y - 2 = 0
Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution,
(ii) only two solutions,
(iii) infinitely many solutions.
The true option is (iii) y = 3x + 5 has infinitely many solutions.
Reason. For every value of x, there is a corresponding value of y and vice-versa.
2x +y = 7
2x + y = 7
⇒ y = 7 – 2x
Put x = 0, we get y = 7 – 2(0) = 7 – 0 = 7
Put x = 1, we gety = 7 – 2(1) = 7 – 2 = 5
Put x = 2, we get y = 7 — 2(2) = 7 – 4 = 3
Put x = 3, we get y = 7 – 2(3) = 7– 6 = 1
∴ Four solutions are (0, 7), (1, 5), (2, 3) and (3, 1).
Sponsor Area
(0) = 9 - 0 = 9
(1)= 9 - 
(-1 = 9 + 
Four solutions are (0, 9) (1, 9 - 
),
x = 4y
Put x = 0, we get y = 
Put x = 4, we get y = 
Put x = 4, we get y = 
Put x = 2, we get y = 
Four solution are (0, 0) (4, 1), (-4, -1) and 
(0, 2)
Put x = 0 and y = 2 in (i)., we get
x - 2y = 0 -2(2) = - 4, which is not 4
(0, 2) is not a solution of (1).
(2, 0)
Put x = 2 and y = 0 in (i)., we get
x - 2y = 2 -2(0) = 2 - 0 = 2, which is not 4
(2, 0) is not a solution of (1).
(4, 0)
Put x = 4 and y = 0 in (i)., we get
x - 2y = 4 -2(0) = 4, which is not 4
(4, 0) is a solution of (1).
, y =
in(1), we get
which is not 4.
is not a solution of (1).(1, 1)
Put x = 1 and y = 1 in (i), we get
x - 2y = 1 -2(1) = 1 - 2 = - 1, which is not 4.
(1, 1) is not a solution of (1).
If x = 2, y = 1 is a solution of the equation 2x + 3y = k, then these values will satisfy the equation. So, putting x = 2 and y = 1 in the equation, we get
2(2) + 3(1) = k
⇒ 4 + 3 = k
⇒ k = 7.
Find at least 3 solutions for the following linear equation in two variables:
2x + 5y = 13
2x + 5y = 13
5y = 13 - 2x 
Put x = 0, then 
Put x = 1, then 
Put x = 2, then 
Put x = 3, then 
are the solutions of equation 2x + 5y = 13.
Find solutions of the form x = a, y = 0 and x = 0, y = b for the following pairs of equations. Do they have any common such solution?
3x + 2y = 6 and 5x + 2y = 10
3x + 2y = 6
Put y = 0, we get 3x + 2(0) = 6
3x = 6
(0, 3) is a solution
5x + 2y = 10
Put y = 0, we get 5x + 2(0) = 10
5x = 10
(0, 5) is a solution.
The given equations have a common solution (2, 0)
x = 3y
x - 3y = 0
1(x) + (- 3)y + (0) = 0
Comparing with ax + by + c = 0, we get
a = 1
b = -3
c = 0
Now, x = 3y
Put x = 0, then y = 
Put x = 3, then y = 
Hence, (0,0) and (3,1) are the two solutions of the equation x = 3y.
- 4m + 27 - 3 = 0
- 4m + 24 = 0
4m = 24
2y = 8 - x
y = 
This expresses y in terms of x.
This line will intersect x-axis at the point for while y = 0. So, put y = 0 in (1), we get
x + 2(0) = 8
⇒ x = 8
Hence, line (1) intersects x-axis at the point (8, 0).
If x = –2, y = 6 is solution of equation
3ax + 2by = 6, then
3a(-2) + 2b(6) = 6
-6a + 12b = 6
-a + 2b = 1 ...(1)
| dividing throughout by 6
Also
2(a - 1) + 2(3b - 4) = 4
2a - 2 + 6b -8 = 4
2a + 6b = 14
a + 3b = 7 ....(2)
| Dividing throughout by 2
Adding (1) and (2), we get
5b = 8 
b = 
Putting b = 
in (1), we get
When x = 2, 
When y = -4, -4 = 
When y = 8, 8 = 
x = 6
When x = -10, y = 
When x = 1, y = 
When y = 
x = -1
Hence, the completed table is as follows:
For intersection with x-axis, put y = 0
Hence, the point of intersection is 
For intersection with y-axis, put x = 0
y = -1
Hence, the intersection is (0, -1)
cuts the x-axis and the y-axis.
Hence the point of intersection with x-axis is (4, 0).
For intersection with y-axis, put x = 0
Hence the point of intersection with y-axis is (0, 6).
times its ordinate.
2x + 5y = 20
Hence the required point is (5, 2)
Find four different solutions of the equation
x + 2y = 6.
Solution not provided.
Ans. (2, 2), (0, 3), (6, 0) and (4, 1).
Find two solutions for each of the following equations:
(i) 2x – 3y = 12 (ii) 2x – 5y = 0
(iii) 3y – 4 = 0.
Solution not provided.
Ans. (i) (0, - 4) and (6, 0)
(ii) (0, 0) and 
(iii) 
Find the value of a so that the following equation may have x = 1, y = 1 as a solution:
3x + ay = 6
Solution not provided.
Ans. 3
Write the equation 
in the form of ax + by + c =0. Check whether (0, 1) and 
are the solutions of the equation.
Solution not provided.
Ans. 
Solution not provided.
Ans. (0, 3) (4, 6), (-4, 0)
Sponsor Area
Solution not provided.
Ans. (0, 4) (4, 7), (8, 10)
Solution not provided.
Ans. a = 1; x - 2y - 10 = 0
Solution not provided.
Ans. k = 7; 
Solution not provided.
Ans. (0, 7), (1, 5), (2, 3), (3, 1), (4, -1), (5, - 3)
Solution not provided.
Ans. y = 9 - 3x; Yes; No
Solution not provided.
Ans. a = 2; (0, 5), (1, 7)
Solution not provided.
Ans. 
Solution not provided.
Ans. 3x - 5y - 4 = 0; no; Yes
Solution not provided.
Ans. a = 4; (0, 2), (1, 1), (2, 0), (3, -1)
Solution not provided.
Ans. 2x + y = 889; 389
Solution not provided.
Ans. Yes; No; No
If 
satisfy the liner equation 3x + ky = 4
, find the value of k, Can there be more than one value of k?
Solution not provided.
Ans. K = -2; Yes
x + y = 4
x + y = 4
⇒ y = 4 – x
Table of solutions
|
X |
1 |
2 |
|
y |
3 |
2 |
We plot the points (1, 3) and (2, 2) on the graph paper and join the same by a ruler to get the line which is the graph of the equation x + y = 4.
x – y = 2
x – y = 2
⇒ y = x – 2
Table of solutions
|
X |
2 |
3 |
|
y |
0 |
1 |
We plot the points (2, 0) and (3, 1) on the graph paper and join the same by a ruler to get the line which is the graph of the equation x – y = 2.
y = 3x
y =3x
Table of solutions
|
X |
0 |
1 |
|
y |
0 |
3 |
We plot the points (0, 0) and (1, 3) on the graph paper and join the same by a ruler to get the line which is the graph of the equation y = 3x.
3 = 2x +y
3 = 2x + y
⇒ y = 3 – 2x
Table of solutions
|
X |
1 |
0 |
|
y |
1 |
3 |
We plot the points (0, 3) and (1, 1) on the graph paper and join the same by a ruler to get the line which is the graph of the equation 3 = 2x + y.
The equations of two lines passing through (2, 14) can be taken as
x + y= 16
and 7x – y = 0.
There are infinitely many such lines because through a point an infinite number of lines can be drawn.
If the point (3, 4) lies on the graph of the equation
3y = ax + 7, then
3(4) = a(3) + 7
12 = 3a + 7
3a = 12 - 7
3a = 5
Total distance covered = x km
Total fare = Rs y
Fare for the first kilometre = Rs 8
Subsequent distance = (x – 1) km
Fare for the subsequent distance = Rs 5(x – 1)
According to the question,
y = 8 + 5 (x – 1)
⇒ y = 8 + 5x – 5
⇒ y = 5x + 3
Table of solutions
|
x |
0 |
1 |
|
y |
3 |
8 |
We plot the points (0, 3) and (1, 8) on the graph paper and join the same by a ruler to get the line which is the graph of the equation y = 5x + 3.
From the choices given below, choose the equation whose graphs are given in Fig. (1) and Fig. (2).
For Fig. 1 For Fig. 2
(i) y = x (i) y = x + 2
(ii) x + y = 0 (ii) y = x – 2
(iii) y = 2x (iii) y = – x + 2
(iv) 2 + 3y = 7x (iv) x + 2y = 6
Fig. 1
Fig. 2
For Fig. (1). The correct equation is
(ii) x + y = 0.
For Fig. (2). The correct equation is
(iii) y = – x + 2.
If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is
(i) 2 units (ii) 0 unit.
Let the work done by the constant force be y units and the distance travelled by the body be x units.
Constant force = 5 units
We know that
Work done = Force x Displacement
⇒ y = 5x
Table of solutions
|
X |
0 |
1 |
|
y |
0 |
5 |
We plot the points (0, 0) and (1, 5) on the graph paper and join the same by a ruler to get the line which is the graph of the equation y = 5x.
(i) Let A → (2, 0), Through A, draw a line parallel to OY to intersect the graph of the equation y = 5x at B. Through B, draw a line parallel to OX to intersect OY at C. Then,
C → (0, 10)
∴ Work done when the distance travelled by the body is 2 units = 10 units.
(ii) Clearly y = 0 when x = 0. So, the work done when the distance travelled by the body is 0 units is 0 units.
Let the contributions of Yamini and Fatima be Rs x and Rs y respectively.
Then according to the question
x + y = 100
This is the linear equation which the given data satisfies.
Now, x + y = 100
⇒ y = 100 – x
Table of solutions
|
X |
0 |
50 |
|
y |
100 |
50 |
We plot the points (0, 100) and (50, 50) on the graph paper and join the same by a ruler to get the line which is the graph of the equation x + y = 100.
Table of solutions
|
c |
0 |
5 |
|
F |
32 |
41 |
We plot the points (0, 32) and (5, 41) on the graph paper and join the same by a ruler to get the line which is the graph of the equation
Required temperature = 860F
When F = 95, then
Required temperature = 350C
When C = 0, Then 
Required temperature = 320F
When F = 0, then
Required temperature = 
When C = 0, then]
Numerical value of required temperature = -40.
3x + y = 8
⇒ y = 8 – 3x
Table of solutions
|
x |
0 |
3 |
|
y |
8 |
-1 |
We plot the points (0, 8) and (3, – 1) on the graph paper and join the same by a ruler to get the line which is the graph of the equation 3jc + y = 8.
∵ The point (2, 2) lies on the graph.
∴ x = 2, y = 2 is a solution of the given equation.
Let the force applied on a body be y units and the acceleration produced in the body be x units. Then,
y ∝ x
⇒ y = kx
Here, k = 5 units |Given
∴ y = 5x ...(1)
(1) is an equation to express the situation. Table of solutions
|
X |
0 |
1 |
|
y |
0 |
5 |
We plot the points (0, 0) and (1, 5) on a graph paper and join the same by a ruler to get the line which is the graph of the equation y = 5x
2x – y = 4
⇒ y = 2x – 4
Table of solutions
|
X |
2 |
3 |
|
y |
0 |
2 |
We plot the points (2, 0) and (3, 2) on a graph paper and join the same by a ruler to get the line which is the graph of the equation 2x – y = 4
x + y = 2
⇒ y = 2 – x
Table of solutions
|
X |
2 |
0 |
|
y |
0 |
2 |
We plot the points (2, 0) and (0, 2) on the same graph paper and join the same by a ruler to get the line which is the graph of the equation x + y = 2.
From graph, we see that the coordinates of the vertices of the triangle are (2, 0), (0, 2) and (0, –4). The triangle has been shaded.
Two lines passing through (1, 2) are
x + y = 3 ...(1)
and y = 2x ...(2)
Infinitely many more such lines can be found because the general equation of a line is ax + by + c = 0. For a given point (x, y) through which the line passes and for an arbitrary pair of values of a and b, c can be determined so as to satisfy ax + by + c = 0. This holds good for each given point and each arbitrary pair of values of a and b. Hence, infinitely many lines can be found passing through a given point.
The given equation is
4x + 3y = 12
3y = 12 - 4x
Table of solutions
|
X |
3 |
0 |
|
y |
0 |
4 |
We plot the points (3, 0) and (0, 4) on a graph paper and join the same by a ruler to get the line which is the graph of the equation 4x + 3y = 12.
x + y = 3 ...(1)
⇒ y = 3 – x
Table of solutions
|
X |
0 |
3 |
|
y |
3 |
0 |
We plot the points (0, 3) and (3, 0) on a graph paper and join these points by a ruler to get the line which is the graph of the equation x + y = 3.
2y = 8 - 2x
y = 4 - xTable of solutions
|
X |
0 |
4 |
|
y |
4 |
0 |
We plot the points (0, 4) and (4, 0) on the same graph paper and join these points by a ruler to get the line which is the graph of the equation 2x + 2y = 8.
From graph, we see that the lines represented by the equations (1) and (2) are parallel.
Draw graph of the following linear equations on the same axes
(i) x + y = 3
(ii) 3x – 2y = 4
Also shade the region formed by their graphs and y-axis.
(i) x + y = 3 ...(1)
⇒ y = 3 – x
Table of solutions
|
X |
0 |
3 |
|
y |
3 |
0 |
We plot the points (0, 3) and (3, 0) on a graph paper and join the same by a ruler to get the line, which is the graph of the equation x + y = 3.
(ii) 3x - 2y = 4 ....(2)
2y = 3x - 4
Table of solutions
|
X |
2 |
4 |
|
y |
1 |
4 |
We plot the points (2, 1) and (4, 4) on the same graph paper on the same axes and join the same by a ruler to get the line which is the graph of the equation 3x – 2y = 4.
The region formed by the graphs of the given linear equations and y-axis has been shaded.
We have
y = 4x + 3 Table of solutions
|
X |
0 |
1 |
|
y |
3 |
7 |
We plot the points (0, 3) and (1, 7) on a graph paper and join the same by a ruler to get the line, which is the graph of the equation y = 4 x + 3.
This gives the work wages graph of the given equation.
Now, on x-axis, take a point P(4, 0). From P draw a line parallel to y-axis intersecting the work wage graph at Q. From Q, draw a line parallel to x-axis to intersect the y-axis at R. We see that R is (0,19).
Check from the graph that (7, 5) is a solution of the linear equation.
Table of solutions
|
x |
1 |
4 |
|
y |
1 |
3 |
We plot the points (1,1) and (4, 3) on a graph paper and join the same by a ruler to get the line which is the graph of the equation 
From graph, we see that the point (7, 5) lies on the graph, so it is a solution of the linear equation.
2x + y = 8
⇒ y = 8 – 2x
Table of solutions
|
x |
4 |
0 |
|
y |
0 |
8 |
We plot the points (4, 0) and (0, 8) on a graph paper and join the same by a ruler to get the line which is the graph of the equation 2x + y = 8.
From graph, we see that this line intersects the x-axis at the point (4, 0) and the y-axis at the point (0, 8).
Let us represent time (in hour) by x and distance (in km) by y. Then, we have y = 80 x
Table of solutions
|
X |
1 |
2 |
3 |
4 |
|
y, |
80 |
160 |
240 |
320 |
We plot the points (1, 80). (2, 160), (3, 240) and (4,320) on a graph paper and join these points by a ruler to get the line which is the graph of the equation
y = 80x
According to the question, the linear equation for the given problem is
⇒ y = 500 + 20x ...(1)
Table of solutions
|
X |
0 |
5 |
|
y |
500 |
600 |
We plot the points (0, 500) and (5, 600) on a graph paper and join the same by a ruler to get the line which is the graph of the equation (1).
4x + 3y = 36 ....(1)
3y = 36 - 4x
Table of solutions
|
X |
0 |
9 |
|
y |
12 |
0 |
We plot the points (0, 12) and (9,0) on a graph paper and join the same by a ruler to get the line which is the graph of the equation (1).
Mark a point P(3, 0) on x-axis. From P draw PQ || y-axis to intersect the graph at Q. From Q, draw QR || x-axis to intersect the y-axis at R(0, 8).
Hence, y = 8 when x = 3.
Again, mark a point A(0, 6) on y-axis. From A draw AB II x-axis to intersect the graph at B. From B, draw BC || y-axis to intersect x-axis at 
.
Hence 
when y = 6
Sponsor Area
Table of solutions
|
x |
0 |
3 |
|
y |
0 |
4 |
We plot the points (0, 0) and (3, 4) on a graph paper and join the same by a ruler to get the line which is the graph of the equation 4x = 3y.
square units.
2x + 3y = -5
3y = - 5 - 2x
Table of solutions
|
X |
–1 |
2 |
|
y |
–1 |
–3 |
We plot the points (–1, –1) and (1, –3) on a graph paper and join the same by a ruler to get the line which is the graph of the equation 2x + 3y = –5.
x + y = –1
⇒ y = – 1 – x
Table of solutions
|
X |
0 |
1 |
|
y |
–1 |
–2 |
We plot the points (0, –1) and (1, –2) on the same graph paper and join the same by a ruler to get the line which is the graph of the equation x + y = –1.
From graph, we see that the coordinates of the point of intersection of the two lines are (2, –3).
Given the point (1,2), can you give the equation of a line on which it lies? How many such equations are there?
Solution not provided.
Ans. x + y = 3, y – x = 1, y = 2x, etc.; Infinitely many.
Solution not provided.
Ans. y = kx where k is a constant.
.(i) If the temperature is 40°C, what is the temperature in Fahrenheit?
(ii) If the temperature is 95°F, what is the temperature in Celcius?
(iii) Is there a temperature which is numerically of same value in Fahrenheit and Celcius? If yes, find the temperature?
Solution not provided.
Ans. (i) 104°F (ii) 35°C (iii) –40°
Solution not provided.
Ans. y = 8 + 5(x – 1) ⇒ y = 5x + 3; र 318
Solution not provided.
Ans. m = 7
Solution not provided.
Ans. x + y = 3; infinitely many;x + y – 3 = 0m = 7
Solution not provided.
Ans. No
Solution not provided.
Ans. x + y = 7; y = x + 1; infinitely many
Solution not provided.
Ans. (4,0); (0,6)
Solution not provided.
Ans. y = 10 + 6(x – 1) ⇒ y = 4 + 6x
Solution not provided.
Ans. x + y = 200 15. (i)1 (ii) 1
Solution not provided.
Ans. y = 3x 17. (i) –4 (ii) 3
Express x in terms of y, it is being given that 7x – 3y = 15. Check if the line represented by the equation intersects the y-axis at y = –5.
Solution not provided.
Ans. 
Solution not provided.
Ans. Rectangle; (0, 0), (5, 0), (5, –3), (0, –3)
Solution not provided.
Ans. 4
Solution not provided.
Ans. 
square units.
Solution not provided.
Ans. h = 0, k = 5
The parking charges of acaron New Delhi Railway Station for first two hours is र 50/- and र 10/- for subsequent hours. Write down an equation and draw the graph of this data. Read the charges from the graph:
(i) for one hour
(ii) for three hours
(iii) for six hours.
Solution not provided.
Ans. y = 50 + 10(x – 2) where x > 2
⇒ y = 30 + 10x where x > 2
(i) र 50 (ii) र 60 (iii) र 90
Give the geometric representations of y = 3 as an equation
in one variable
In one variable
The representation of y = 3 on the number line is as shown below:
Give the geometric representations of y = 3 as an equation
in two variables.
In two variables
y = 3
⇒ 0.x + 1.y = 3
It is a linear equation in two variables x and y. This is represented by a line. All the values of x are permissible because O.x is always 0. However, y must satisfy the relation y = 3. Hence, two solutions of the given equation are x = 0, y = 3 and x = 2, y = 3.
Thus the graph AB is a line parallel to the x-axis at a distance of 3 units above it.
The given equation is
2x + 9 = 0
In one variable
2x + 9 = 0
⇒ 2x = – 9
The representation of 2x + 9 = 0 on the number line is as shown below:
In two variables
2x + 9 = 0
⇒ 2x + Oy + 9 = 0
It is a linear equation in two variables x and y. This is represented by a line. All the values of y are permissible because Oy is always 0. However, x must satisfy the relation
2x + 9 = 0, i.e. 
Hence, two solutions of the given equation are Hence,
The graph AB is a line parallel to the y-axis and at a distance of 
units to the left of origin O.
Solve for 
. What will be the graph of this equation?
Solution not provided.
Ans. x = 1 5. x = 4
Solve the equation 3(x + 2) = 2(2x – 1) and represent the solution:
(i) on the number line
(ii) in the Cartesian plane.
Solution not provided.
Ans. x = 3
Solution not provided.
Solution not provided.
Solve 4x – 7 = 9. Represent the solution (i) on the number line
(ii) in the Cartesian plane.
Solution not provided.
Give the geometric interpretation of 7x + 6 = 2x – 4 as an equation:
(i) in one variable
(ii) in two variables.
Solution not provided.
Give the geometrical representation of the equation 3x + 15 = 0 as an equation:
(i) in one variable
(ii) in two variables
Solution not provided.
Solve 5x – 2 = 3x – 8 and represent the solution.
(i) on a number line
(ii) in the Cartesian plane.
Solution not provided.
Solve the equation 3x + 4 = 7 + 2x for x and represent the solution
(i) On a number line
(ii) In the Cartesian plane
Solution not provided.
by = c
A.
ax + by + c = 0A.
–1, 2, –7D.
infinitely many solutionsD.
infinitely many solutionsSponsor Area
B.
does not pass through originWhere does the line 2x + 3y = 6 cut x-axis?
A.
at (3, 0)C.
infinitely manyA.
parallel to x-axis at a distance 6 units from the originD.
infinitely many solutionsC.
x + y = 0D.
parallel to x = –4
D.
on the y-axis2 . x + 3 . y = 0
2 . x + 0 . y = 3
1 . x + 
B.
2 . x + 0 . y = 3
3x -7y = 10
y - 2x = 3
8y - 6x = 4
C.
8y - 6x = 4
ax + b = 0 where a, b are real numbers and a ≠ b
A.
ax + by + c = 0, where a, b, c are real numbers and a, b ≠ 0C.
line that passes through the originSolution not provided.
Ans. x + y + 1 = 0; 2x + y = 1; 3x + y = 3
Solution not provided.
Ans. (0, 0) (1, - 2), (2, -4)
Solution not provided.
Ans. y = 5 + 2(x -1) 
y = 3 + 2x
Consider the point A(–2, 3)
(i) How many lines can be drawn passing through A?
(ii) Give equation of any two lines passing through A.
(iii) Without actually drawing the graph, find a point, other than A, on each of these two lines.
Solution not provided.
Ans. (i) infinitely many (ii) x +y + 1 = 0, -x + y = 5
(iii) (0, 1), (0, 5)
Solution not provided.
Ans. (0, 0)
Solution not provided.
Ans. (2, 0), (0, 6)
Solution not provided.
Ans. y = 3x; Yes
Solution not provided.
Ans. (1, 1)
Solution not provided.
Ans. k = 7
Solution not provided.
Ans. x + y + 1 = 0, 2y + x = 2, 3y + x =9; infintely many
Solution not provided.
Ans. y = 2x + 6
Solution not provided.
Ans. y =4x; Rs. 44
Determine the co-ordinates of a point on the graph of 5x – y = 12 whose
(a) ordinate is twice that of the abscissa
(b) abscissa and ordinate are in the ratio 3 : 2.
Solution not provided.
Ans. (4, 2) (b) 
A rectangular field has to be cut out and its boundary marked with fencing with a given wire of length 100 m.
(a) Represent the above situation using a linear equation
(b) Also plot its graph
Solution not provided.
Ans. (a) d + y = 50
(i) Only one line can pass through a single point.
(ii) There are an infinite number of lines which pass through two distinct points.
(iii) A terminated line can be produced indefinitely on both the sides.
(iv) If two circles are equal, then their radii are equal.
(v) In figure, if AB = PQ and PQ = XY, then AB = XY.
(i) False. This can be seen usually.
(ii) False. This contradicts Axiom 5.1.
(iii) True. Postulate 2.
(iv) True. If we superimpose the region bounded by one circle on the other, then they coincide. So, their centres and boundaries coincide therefore, their radii will coincide.
(v) True. The first Axiom of Euclid.
Give a definition for each of the following terms. Are there other terms that need to be defined first ? What are they, and how might you define them?
(i) parallel lines (ii) perpendicular lines
(iii) line segment (iv) radius of a circle
(v) square.
(i) Parallel lines. Lines which do not intersect anywhere are called parallel lines.
(ii) Perpendicular lines. Two lines which are at a right angle to each other are called perpendicular lines.
(iii) Line segment. It is a terminated line.
(iv) Radius. The length of the line-segment joining the centre of a circle to any point on its circumference is called its radius.
(v) Square. A quadrilateral with all the four sides equal and all the four angles of measure 90° each is called a square.
Consider two ‘postulates’ given below:
(i) Given any two distinct points A and B, there exists a third point C which is in between A and B.
(ii) There exist at least three points that are not on the same line.\
Do these postulates contain any undefined terms ? Are these postulates consistent? Do they follow from Euclid’s postulates ? Explain.
Yes! These postulates contain two undefined terms: Point and Line.
Yes! These postulates are consistent because they deal with two different situations (i) say that given two points A and B, there is a point C lying on the line in between them, (ii) say that given A and B, we can take C not lying on the line through A and B. These ‘postulates’ do not follow from Euclid’s postulates however, they follow from Axiom 5.1.
AB. Explain by drawing the figure.
2AC = AB
AC = BD | Given ...(1)
AC = AB + BC
| Point B lies between A and C ...(2)
BD = BC + CD
| Point C lies between B and D ...(3)
Substituting (2) and (3) in (1), we get
AB + BC = BC + CD ⇒ AB = CD.
| Subtracting equals from equals
AC = XD | Given
⇒ 2AC = 2XD
| Things which are double of the same things are equal to one another
⇒ AB = XY
| ∵ C is the mid-point of AB and D is the mid-point of XY
∠2 = ∠4 ...(1) | Given
∠1 = ∠4 ...(2) | Given
∠3 = ∠2 ...(3) | Given
From (1),
∠4 = ∠2 ...(4)
From (3) and (4),
∠3 = ∠4 ...(5)
| Things which are equal to the same thing are equal to one another
From (2) and (4),
∠1 = ∠3
|Things which are equal to the same thing are equal to one another
Solution not provided.
Solution not provided.
Solution not provided.
If a straight line l falls on two straight lines m and n such that sum of the interior angles on one side of l is two right angles, then by Euclid’s fifth postulate the lines m and n will not meet on this side of I. Next, we know that the sum of the interior angles on the other side of line l will also be two right angles.
Therefore, they will not meet on the other side also. So, the lines m and n never meet and are, therefore parallel.
C.
Two lines drawn in a plane always intersect at a pointC.
Two distinct lines cannot have more than one point in common.A.
Only one line can pass through a single point.D.
Infinitely manyC.
straight lineA.
equal to anotherSponsor Area
Sponsor Area
then the value of 
and y = 1, then value of x is
