NCERT Exemplar: Pair of Linear Equations in Two Variables Notes | EduRev

Mathematics (Maths) Class 10

Class 10 : NCERT Exemplar: Pair of Linear Equations in Two Variables Notes | EduRev

The document NCERT Exemplar: Pair of Linear Equations in Two Variables Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
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Exercise 3.1

Choose the correct answer from the given four options:
1. Graphically, the pair of equations
6x – 3y + 10 = 0
2x – y + 9 = 0
represents two lines which are
(a) Intersecting at exactly one point.
(b) Intersecting at exactly two points.
(c) Coincident
(d) parallel.
Ans:
(d)
Explanation:
The given equations are,
6x - 3y + 10 = 0
dividing by 3
⇒ 2x - y + 10/3 = 0… (i)
And 2x - y + 9 = 0…(ii)
Table for 2x - y + 10/3 = 0,

x

0

-5/3

y

10/3

0

Table for 2x - y + 9 = 0

x

0

-9/2

y

9

0

NCERT Exemplar: Pair of Linear Equations in Two Variables Notes | EduRev

Hence, the pair of equations represents two parallel lines.

2. The pair of equations x + 2y + 5 = 0 and –3x – 6y + 1 = 0 have
(a) a unique solution
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution
Ans:
(d)
Explanation:
The equations are:
+ 2+ 5 = 0
–3– 6y + 1 = 0
a= 1; b1 = 2; c= 5
a= -3; b2 = -6; c= 1
a1/a= -1/3
b1/b2 = -2/6 = -1/3
c1/c= 5/1 = 5
Here,
a1/a= b1/b2 ≠ c1/c2
Therefore, the pair of equation has no solution.

3. If a pair of linear equations is consistent, then the lines will be
(a) parallel 
(b) always coincident
(c) intersecting or coincident 
(d) always intersecting
Ans: (c)
Explanation:
Condition for a pair of linear equations to be consistent are:
Intersecting lines having unique solution,
a1/a≠ b1/b2
Coincident or dependent
a1/a= b1/b2 = c1/c2

4. The pair of equations y = 0 and y = –7 has
(a) one solution
(b) two solutions
(c) infinitely many solutions
(d) no solution

Ans: (d)
Explanation:
The given pair of equations are y = 0 and y = – 7.

NCERT Exemplar: Pair of Linear Equations in Two Variables Notes | EduRevGraphically, both lines are parallel and have no solution

5. The pair of equations x = a and y = b graphically represents lines which are
(a) parallel
(b) intersecting at (b, a)
(c) coincident
(d) intersecting at (a, b)
Ans:
(d)
Explanation: 
Graphically in every condition,
a, b>>0
a, b< 0
a>0, b< 0
a<0, b>0 but a = b≠ 0.
The pair of equations x = a and y = b graphically represents lines which are intersecting
at (a, b).
NCERT Exemplar: Pair of Linear Equations in Two Variables Notes | EduRev

Hence, the cases two lines intersect at (a, b).


Exercise 3.2

1. Do the following pair of linear equations have no solution? Justify your answer.
(i) 2x + 4y = 3
12y + 6x = 6
(ii) x = 2y
y = 2x
(iii) 3x + y – 3 = 0
2x + 2/3y = 2
Solution:

The Condition for no solution = a1/a= b1/b2 ≠ c1/c2 (parallel lines)
(i) Yes.
Given pair of equations are,
2x + 4y – 3 = 0 and 6x + 12y – 6 = 0
Comparing the equations with ax+ by +c = 0;
We get,
a1 = 2, b1 = 4, c1 = – 3;
a2 = 6, b2 = 12, c2 = – 6;
a1 /a2 = 2/6 = 1/3
b1 /b2 = 4/12 = 1/3
c1 /c2 = – 3/ – 6 = ½
Here, a1/a2 = b1/b≠ c1/c2, i.e parallel lines
Hence, the given pair of linear equations has no solution.
(ii) No.
Given pair of equations,
x = 2y or x – 2y = 0
y = 2x or 2x – y = 0;
Comparing the equations with ax+ by +c = 0;
We get,
a1 = 1, b1 = – 2, c1 = 0;
a2 = 2, b2 = – 1, c2 = 0;
a1 /a2 = ½
b1 /b2 = -2/-1 = 2
Here, a1/a2 ≠ b1/b2.
Hence, the given pair of linear equations has unique solution.
(iii) No.
Given pair of equations,
3x + y – 3 = 0
2x + 2/3 y = 2
Comparing the equations with ax+ by +c = 0;
We get,
a1 = 3, b1 = 1, c1 = – 3;
a2 = 2, b2 = 2/3, c2 = – 2;
a1 /a2 = 2/6 = 3/2
b1 /b2 = 4/12 = 3/2
c1 /c2 = – 3/-2 = 3/2
Here, a1/a2 = b1/b2 = c1/c2, i.e coincident lines

2. Do the following equations represent a pair of coincident lines? Justify your answer.
(i) 3x + 1/7y = 3
7x + 3y = 7
(ii) –2x – 3y = 1
6y + 4x = – 2
(iii) x/2 + y + 2/5 = 0
4x + 8y + 5/16 = 0
Solution:

Condition for coincident lines,
a1/a2 = b1/b2 = c1/c2;
(i) No.
Given pair of linear equations are:
3x + 1/7y = 3
7x + 3y = 7
Comparing the above equations with ax + by + c = 0;
Here, a1 = 3, b1 = 1/7, c1 = – 3;
And a2 = 7, b2 = 3, c2 = – 7;
a1 /a2 = 3/7
b1 /b2 = 1/21
c1 /c2 = – 3/ – 7 = 3/7
Here, a1/a2 ≠ b1/b2.
Hence, the given pair of linear equations has unique solution.
(ii) Yes,
Given pair of linear equations.
– 2x – 3y – 1 = 0 and 4x + 6y + 2 = 0;
Comparing the above equations with ax + by + c = 0;
Here, a1 = – 2, b1 = – 3, c1 = – 1;
And a2 = 4, b2 = 6, c2 = 2;
a1 /a2 = – 2/4 = – ½
b1 /b2 = – 3/6 = – ½
c1 /c2 = – ½
Here, a1/a2 = b1/b2 = c1/c2, i.e. coincident lines
Hence, the given pair of linear equations is coincident.
(iii) No,
Given pair of linear equations are
x/2 + y + 2/5 = 0
4x + 8y + 5/16 = 0
Comparing the above equations with ax + by + c = 0;
Here, a1 = ½, b1 = 1, c1 = 2/5;
And a2 = 4, b2 = 8, c2 = 5/16;
a1 /a2 = 1/8
b1 /b2 = 1/8
c1 /c2 = 32/25
Here, a1/a2 = b1/b≠ c1/c2, i.e. parallel lines
Hence, the given pair of linear equations has no solution.

3. Are the following pair of linear equations consistent? Justify your answer.
(i) –3x– 4y = 12
4y + 3x = 12
(ii) (3/5)x – y = ½
(1/5)x – 3y= 1/6
(iii) 2ax + by = a
1. ax + 2by – 2a = 0; a, b ≠ 0
(iv) x + 3y = 11
2 (2x + 6y) = 22
Solution:

Conditions for pair of linear equations to be consistent are:
a1/a2 ≠ b1/b2. [unique solution]
a1/a2 = b1/b2 = c1/c2 [coincident or infinitely many solutions]
(i) No.
The given pair of linear equations
– 3x – 4y – 12 = 0 and 4y + 3x – 12 = 0
Comparing the above equations with ax + by + c = 0;
We get,
a1 = – 3, b1 = – 4, c1 = – 12;
a2 = 3, b2 = 4, c2 = – 12;
a1 /a2 = – 3/3 = – 1
b1 /b2 = – 4/4 = – 1
c1 /c2 = – 12/ – 12 = 1
Here, a1/a2 = b1/b≠ c1/c2
Hence, the pair of linear equations has no solution, i.e., inconsistent.
(ii) Yes.
The given pair of linear equations
(3/5)x – y = ½
(1/5)x – 3y= 1/6
Comparing the above equations with ax + by + c = 0;
We get,
a1 = 3/5, b1 = – 1, c1 = – ½;
a2 = 1/5, b2 = 3, c2 = – 1/6;
a1 /a2 = 3
b1 /b2 = – 1/ – 3 = 1/3
c1 /c2 = 3
Here, a1/a2 ≠ b1/b2.
Hence, the given pair of linear equations has unique solution, i.e., consistent.
(iii) Yes.
The given pair of linear equations –
2ax + by –a = 0 and 4ax + 2by – 2a = 0
Comparing the above equations with ax + by + c = 0;
We get,
a1 = 2a, b1 = b, c1 = – a;
a2 = 4a, b2 = 2b, c2 = – 2a;
a1 /a2 = ½
b1 /b2 = ½
c1 /c2 = ½
Here, a1/a2 = b1/b2 = c1/c2
Hence, the given pair of linear equations has infinitely many solutions, i.e., consistent
(iv) No.

The given pair of linear equations
x + 3y = 11 and 2x + 6y = 11
Comparing the above equations with ax + by + c = 0;
We get,
a1 = 1, b1 = 3, c1 = 11
a2 = 2, b2 = 6, c2 = 11
a1 /a2 = ½
b1 /b2 = ½
c1 /c2 = 1
Here, a1/a2 = b1/b≠ c1/c2.
Hence, the given pair of linear equations has no solution.


Exercise 3.3

1. For which value(s) of λ, do the pair of linear equations
λx + y = λ2 and x + λy = 1 have
(i) no solution?
(ii) infinitely many solutions?
(iii) a unique solution?
Solution:

The given pair of linear equations is
λx + y = λ2 and x + λy = 1
a1 = λ, b1= 1, c1 = – λ2
a2 = 1, b= λ, c2= -1
The given equations are;
λ x + y – λ2 = 0
x + λ y – 1 = 0
Comparing the above equations with ax + by + c = 0;
We get,
a1 = λ, b1 = 1, c1 = – λ 2;
a2 = 1, b2 = λ, c2 = – 1;
a1 /a2 = λ/1
b1 /b2 = 1/λ
c1 /c2 = λ2
(i) For no solution,
a1/a2 = b1/b2≠ c1/c2
i.e. λ = 1/ λ ≠ λ2
so, λ 2 = 1;
and λ 2 ≠ λ
Here, we take only λ = – 1,
Since the system of linear equations has no solution.
(ii) For infinitely many solutions,
a1/a2 = b1/b2 = c1/c2
i.e. λ = 1/ λ  = λ2
so λ = 1/ λ gives λ = + 1;
λ = λ 2 gives λ = 1,0;
Hence satisfying both the equations
λ = 1 is the answer.
(iii) For a unique solution,
a1/a2 ≠ b1/b2
so λ ≠1/ λ
hence, λ2 ≠ 1;
λ ≠ + 1;
So, all real values of λ except +1

2. For which value(s) of will the pair of equations
kx + 3= – 3
12+ ky = k
have no solution?
Solution:
The given pair of linear equations is
kx + 3y = k – 3 …(i)
12x + ky = k …(ii)
On comparing the equations (i) and (ii) with ax + by = c = 0,
We get,
a1 = k, b1 = 3, c1 = -(k – 3)
a2 = 12, b2 = k, c2 = – k
Then,
a1 /a2 = k/12
b1 /b2 = 3/k
c1 /c2 = (k-3)/k
For no solution of the pair of linear equations,
a1/a2 = b1/b2≠ c1/c2
k/12 = 3/k ≠ (k-3)/k
Taking first two parts, we get
k/12 = 3/k
k2 = 36
k = + 6
Taking last two parts, we get
3/k ≠ (k-3)/k
3k ≠ k(k – 3)
k2 – 6k ≠ 0
so, k ≠ 0, 6
Therefore, value of k for which the given pair of linear equations has no solution is k = – 6.

3. For which values of and b, will the following pair of linear equations have infinitely many solutions?
+ 2= 1
(b)+ (+ b)= + – 2
Solution:
The given pair of linear equations are:
x + 2y = 1 …(i)
(a - b)x + (a + b)y = a + b – 2 …(ii)
On comparing with ax + by = c = 0 we get
a1 = 1, b1 = 2, c1 = – 1
a2 = (a – b), b2 = (a + b), c2 = – (a + b – 2)
a1 /a2 = 1/(a - b)
b1 /b2 = 2/(a + b)
c1 /c2 = 1/(a + b - 2)
For infinitely many solutions of the, pair of linear equations,
a1/a2 = b1/b2=c1/c2(coincident lines)
so, 1/(a - b) = 2/ (a + b) = 1/(a + b - 2)
Taking first two parts,
1/(a-b) = 2/ (a+b)
a + b = 2(a – b)
a = 3b …(iii)
Taking last two parts,
2/ (a + b) = 1/(a + b - 2)
2(a + b – 2) = (a + b)
a + b = 4 …(iv)
Now, put the value of a from Eq. (iii) in Eq. (iv), we get
3b + b = 4
4b = 4
b = 1
Put the value of b in Eq. (iii), we get
a = 3
So, the values (a, b) = (3, 1) satisfies all the parts. Hence, required values of a and b are 3 and 1 respectively for which the given pair of linear equations has infinitely many solutions.

4. Find the value(s) of p in (i) to (iv) and p and q in (v) for the following pair of equations:
(i) 3x – y – 5 = 0 and 6x – 2y – p = 0, if the lines represented by these equations are parallel.
Solution:
Given pair of linear equations is
3x – y – 5 = 0 …(i)
6x – 2y – p = 0 …(ii)
On comparing with ax + by + c = 0 we get
We get,
a1 = 3, b1 = – 1, c1 = – 5;
a2 = 6, b2 = – 2, c2 = – p;
a1 /a2 = 3/6 = ½
b1 /b2 = ½
c1 /c2 = 5/p
Since, the lines represented by these equations are parallel, then
a1/a2 = b1/b≠ c1/c2
Taking last two parts, we get ½ ≠ 5/p
So, p ≠ 10
Hence, the given pair of linear equations are parallel for all real values of p except 10.

(ii) – + py = 1 and px = 1, if the pair of equations has no solution.
Solution:
Given pair of linear equations is
– x + py = 1 …(i)
px – y – 1 = 0 …(ii)
On comparing with ax + by + c = 0,
We get,
a1 = -1, b1 = p, c1 =- 1;
a2 = p, b2 = – 1, c2 =- 1;
a1 /a2 = -1/p
b1 /b2 = – p
c1 /c2 = 1
Since, the lines equations has no solution i.e., both lines are parallel to each other.
a1/a2 = b1/b2≠ c1/c2
-1/p = – p ≠ 1
Taking last two parts, we get
p ≠ -1
Taking first two parts, we get
p2 = 1
p = + 1
Hence, the given pair of linear equations has no solution for p = 1.

(iii) – 3+ 5= 7 and 2px – 3= 1, if the lines represented by these equations are intersecting at a unique point.
Solution:
Given, pair of linear equations is
– 3x + 5y = 7
2px – 3y = 1
On comparing with ax + by + c = 0, we get
Here, a1 = -3, b1 = 5, c1 = – 7;
And a2 = 2p, b2 = – 3, c2 = – 1;
a1 /a2 = -3/ 2p
b1 /b2 = – 5/3
c1 /c2 = 7
Since, the lines are intersecting at a unique point i.e., it has a unique solution
a1/a2 ≠ b1/b2
-3/2p ≠ -5/3
p ≠ 9/10
Hence, the lines represented by these equations are intersecting at a unique point for all real values of p except 9/10

(iv) 2+ 3– 5 = 0 and px – 6– 8 = 0, if the pair of equations has a unique solution.
Solution:
Given, pair of linear equations is
2x + 3y – 5 = 0
px – 6y – 8 = 0
On comparing with ax + by + c = 0 we get
Here, a1 = 2, b1 = 3, c1 = – 5;
And a2 = p, b2 = – 6, c2 = – 8;
a1 /a2 = 2/p
b1 /b2 = – 3/6 = – ½
c1 /c2 = 5/8
Since, the pair of linear equations has a unique solution.
a1/a2 ≠ b1/b2
so 2/p ≠ – ½
p ≠ – 4
Hence, the pair of linear equations has a unique solution for all values of p except – 4.

(v) 2+ 3= 7 and 2px + py = 28 – qy, if the pair of equations have infinitely many solutions.
Solution:
Given pair of linear equations is
2x + 3y = 7
2px + py = 28 – qy
or 2px + (p + q)y – 28 = 0
On comparing with ax + by + c = 0,
We get,
Here, a1 = 2, b1 = 3, c1 = – 7;
And a2 = 2p, b2 = (p + q), c2 = – 28;
a1/a2 = 2/2p
b1/b= 3/ (p + q)
c1/c= ¼
Since, the pair of equations has infinitely many solutions i.e., both lines are coincident.
a1/a2 = b1/b2 = c1/c2
1/p = 3/(p+q) = ¼
Taking first and third parts, we get
p = 4
Again, taking last two parts, we get
3/(p+q) = ¼
p + q = 12
Since p = 4
So, q = 8
Here, we see that the values of p = 4 and q = 8 satisfies all three parts.
Hence, the pair of equations has infinitely many solutions for all values of p = 4 and q = 8.

5. Two straight paths are represented by the equations – 3= 2 and –2+ 6= 5.Check whether the paths cross each other or not.
Solution:
Given linear equations are
x – 3y – 2 = 0 …(i)
-2x + 6y – 5 = 0 …(ii)
On comparing with ax + by c = 0,
We get
a1 = 1, b1 = -3, c1 = - 2;
a2 = -2, b2 = 6, c2 = - 5;
a1/a2 = – ½
b1/b2 = – 3/6 = – ½
c1/c2 = 2/5
i.e., a1/a2 = b1/b2 ≠ c1/c2 [parallel lines]
Hence, two straight paths represented by the given equations never cross each other, because they are parallel to each other.

6. Write a pair of linear equations which has the unique solution = – 1, =3. How many such pairs can you write?
Solution:
Condition for the pair of system to have unique solution
a1/a2 ≠ b1/b2
Let the equations be,
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
Since, x = – 1 and y = 3 is the unique solution of these two equations, then
It must satisfy the equations –
a1(-1) + b1(3) + c1 = 0
– a1 + 3b1 + c1 = 0 …(i)
and a2(- 1) + b2(3) + c2 = 0
– a2 + 3b2 + c2 = 0 …(ii)
Since for the different values of a1, b1, c1 and a2, b2, c2 satisfy the Eqs. (i) and (ii).
NCERT Exemplar: Pair of Linear Equations in Two Variables Notes | EduRev

Hence, infinitely many pairs of linear equations are possible.

7. If 2+ = 23 and 4= 19, find the values of 5– 2and y/x – 2.
Solution:
Given equations are
2x + y = 23 …(i)
4x – y = 19 …(ii)
On adding both equations, we get
6x = 42
So, x = 7
Put the value of x in Eq. (i), we get
2(7) + y = 23
y = 23 – 14
so, y = 9
Hence 5y – 2x = 5(9) – 2(7) = 45 – 14 = 31
y/x – 2 = 9/7 -2 = -5/7
Hence, the values of (5y – 2x) and y/x – 2 are 31 and -5/7 respectively.

8. Find the values of x and y in the following rectangle [see Fig. 3.2].
NCERT Exemplar: Pair of Linear Equations in Two Variables Notes | EduRev

Solution:
Using property of rectangle,
We know that,
Lengths are equal,

i.e., CD = AB

Hence, x + 3y = 13 …(i)

Breadth are equal,

i.e., AD = BC

Hence, 3x + y = 7 …(ii)

On multiplying Eq. (ii) by 3 and then subtracting Eq. (i),

We get,

8x = 8

So, x = 1

On substituting x = 1 in Eq. (i),

We get,

y = 4

Therefore, the required values of x and y are 1 and 4, respectively.


Exercise 3.4

1. Graphically, solve the following pair of equations:
2x + y = 6
2x – y + 2 = 0
Find the ratio of the areas of the two triangles formed by the lines representing these equations with the x-axis and the lines with the y-axis.
Solution: 

Given equations are 2x + y = 6 and 2x – y + 2 = 0
Table for equation 2x + y – 6 = 0, for x = 0, y = 6, for y = 0, x = 3.

x

0

3

y

6

0

Table for equation 2x – y + 2 = 0, for x = 0, y = 2, for y = 0,x = – 1

x

0

-1

y

2

0

Let A1 and A2 represent the areas of triangles ACE and BDE respectively.
NCERT Exemplar: Pair of Linear Equations in Two Variables Notes | EduRev

Let, Area of triangle formed with x -axis = T1
T1 = Area of △ACE = ½ × AC × PE
T1 = ½ × 4 × 4 = 8
And Area of triangle formed with y – axis = T2
T1 = Area of △BDE = ½ × BD × QE
T1 = ½ × 4 × 1 = 2
T1:T2 = 8:2 = 4:1
Hence, the pair of equations intersect graphically at point E(1, 4)
i.e., x = 1 and y = 4.

2. Determine, graphically, the vertices of the triangle formed by the lines
y = x, 3y = x, x + y = 8
Solution: 
Given linear equations are
y = x …(i)
3y = x …(ii)
and x + y = 8 …(iii)
Table for line y = x, 

x

0

1

2

y

0

1

2

Table for line x = 3y,

x

0

3

6

y

0

1

2

Table for line x + y = 8

x

0

4

8

y

8

4

0

Plotting the points A (1, 1), B(2, 2), C (3, 1), D (6, 2), we get the straight lines AB and CD.
Similarly, plotting the point P (0, 8), Q(4, 4) and R(8, 0), we get the straight line PQR.
AB and CD intersects the line PR on Q and D, respectively.
NCERT Exemplar: Pair of Linear Equations in Two Variables Notes | EduRev

So, △OQD is formed by these lines. Hence, the vertices of the △OQD formed by the given lines are O(0, 0), Q(4, 4) and D(6, 2).

3. Draw the graphs of the equations x = 3, x = 5 and 2x – y – 4 = 0. Also find the area of the quadrilateral formed by the lines and the x–axis.
Solution: 
Given equation of lines x = 3, x = 5 and 2x - y - 4 = 0.
Table for line 2x – y – 4 = 0

x

0

2

y

-4

0

Plotting the graph, we get, 

NCERT Exemplar: Pair of Linear Equations in Two Variables Notes | EduRev

From the graph, we get,
AB = OB-OA = 5-3 = 2
AD = 2
BC = 6
Thus, quadrilateral ABCD is a trapezium, then,
Area of Quadrileral ABCD = ½ × (distance between parallel lines) = ½ × (AB) × (AD + BC)
= 8 sq units

4. The cost of 4 pens and 4 pencil boxes is Rs 100. Three times the cost of a pen is Rs 15 more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box.
Solution:
Let the cost of a pen and a pencil box be Rs x and Rs y respectively.
According to the question,
4x + 4y = 100
Or x + y = 25 …(i)
3x = y + 15
Or 3x - y = 15 …(ii)
On adding Equation (i) and (ii), we get,
4x = 40
So, x = 10
Substituting x = 10, in Eq. (i) we get
y = 25 - 10 = 15
Hence, the cost of a pen = Rs. 10
The cost of a pencil box = Rs. 15

5. Determine, algebraically, the vertices of the triangle formed by the lines
3= 3
2– 3= 2
+ 2= 8
Solution:
3x – y = 2 …(i)
2x -3y = 2 …(ii)
x + 2y = 8 …(iii)
Let the equations of the line (i), (ii) and (iii) represent the side of a ∆ABC.
On solving (i) and (ii),
We get,
[First, multiply Eq. (i) by 3 in Eq. (i) and then subtract]
(9x - 3y)-(2x - 3y) = 9 - 2
7x = 7
x = 1
Substituting x=1 in Eq. (i), we get
3 × 1 - y = 3
y = 0
So, the coordinate of point B is (1, 0)
On solving lines (ii) and (iii),
We get,
[First, multiply Eq. (iii) by 2 and then subtract]
(2x + 4y)-(2x-3y) = 16-2
7y = 14
y = 2
Substituting y=2 in Eq. (iii), we get
x + 2 × 2 = 8
x + 4 = 8
x = 4
Hence, the coordinate of point C is (4, 2).
On solving lines (iii) and (i),
We get,
[First, multiply in Eq. (i) by 2 and then add]
(6x - 2y) + (x + 2y) = 6 + 8
7x = 14
x = 2
Substituting x=2 in Eq. (i), we get
3 × 2 – y = 3
y = 3
So, the coordinate of point A is (2, 3).
Hence, the vertices of the ∆ABC formed by the given lines are as follows,
A (2, 3), B (1, 0) and C (4, 2).

6. Ankita travels 14 km to her home partly by rickshaw and partly by bus. She takes half an hour if she travels 2 km by rickshaw, and the remaining distance by bus.
Solution:
Let the speed of the rickshaw and the bus are x and y km/h, respectively.
Now, she has taken time to travel 2 km by rickshaw, t1 = (2/x) hr
Speed = distance/ time
she has taken time to travel remaining distance i.e., (14 – 2) = 12km
By bus t2 = (12/y) hr
By first condition,
t+ t2 = ½ = (2/x) + (12/y) … (i)
Now, she has taken time to travel 4 km by rickshaw, t3 = (4/x) hr
and she has taken time to travel remaining distance i.e., (14 – 4) = 10km, by bus = t4 = (10/y) hr
By second condition,
t3 + t4 = ½ + 9/60 = ½ + 3/20
(4/x) + (10/y) = (13/20) …(ii)
Let (1/x) = u and (1/y) = v
Then Equations. (i) and (ii) becomes
2u + 12v = ½ …(iii)
4u + 10v = 13/20…(iv)
[First, multiply Eq. (iii) by 2 and then subtract]
(4u + 24v) – (4u + 10v) = 1–13/20
14v = 7/20
v = 1/40
Substituting the value of v in Eq. (iii),
2u + 12(1/40) = ½
2u = 2/10
u = 1/10
x = 1/u = 10km/hr
y = 1/v = 40km/hr
Hence, the speed of rickshaw = 10 km/h
And the speed of bus = 40 km/h.

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