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The line represented by x = 7 is parallel to the x-axis. Justify whether the statement is true or not.


Solution:

From the above graph, it is clear that x = 7 is not parallel to the x-axis but it is parallel to the y-axis. Hence the given statement is false.

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For all real values of c, the pair of equations x– 2y = 8, 5x – 10y = c have a unique solution. Justify whether it is true or false.

Solution:
Equation are x – 2y = 8
5x – 10y = c
Here a1 = 1, b1 = –2, c1 = –8
a2 = 5, b2 = –10, c2 = –c

\frac{a_{1}}{a_{2}}= \frac{1}{5};\frac{b_{1}}{b_{2}}= \frac{-2}{-10}= \frac{1}{5};\frac{c_{1}}{c_{2}}= \frac{-8}{-c}= \frac{8}{c}
For unique solution \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}} but here \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}

Hence the given statement is false.

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For the pair of equations \lambdax + 3y = –7 ,2x + 6y = 14to have infinitely many solutions, the value of \lambda should be 1. Is the statement true?Give reasons.

Solution:
Here equations are \lambda x + 3y = –7
2x + 6y = 14
a1 = \lambda , b1 = 3, c1 = 7
a2 = 2, b2 = 6, c2 = –14
The equation have infinitely many solution

\therefore \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}
\frac{\lambda }{2}= \frac{3}{6}\neq \frac{-7}{14}
Here we can see that

\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}
Hence no value of \lambda exist because it is given that equation has infinitely many solutions
Hence the statement is false.

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Are the following pair of linear equations consistent? Justify your answer.

(i) –3x– 4y = 12 ; 4y + 3x = 12

(ii) \frac{3}{5}x-y= \frac{1}{2};\frac{1}{5}x-3y= \frac{1}{6} ;
(iii) 2ax + by = a; 4ax + 2by = 2a ; a, b ≠ 0
(iv) x + 3y = 11 ; 4x + 12y = 22

 

 

Solution: (i)
Equation are : –3x – 4y = 12
4y + 3x = 12
In equation
–3x – 4y – 12 = 0
a1 = –3; b1 = –4 ; c1 = –12
In equation
4y + 3x – 12 = 0
a2 = 3 ; b2 = 4; c2 = –12
\frac{a_{1}}{a_{2}}= \frac{-3}{3}= -1;\frac{b_{1}}{b_{2}}= \frac{-4}{4}= -1,\frac{c_{1}}{c_{2}}\Rightarrow \frac{-12}{-12}= \frac{1}{1}

For consistency either \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}or \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}} but here \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}

Hence the pair of linear equations is not consistent.
Solution: (ii)
Equation are : \frac{3}{5} x-y=\frac{1}{2}

\frac{1}{5} x-3 y=\frac{1}{6}
In equation 

\frac{3}{5} x-y=\frac{1}{2}
a1= \frac{3}{5}; b1 = –1 ; c1 = -\frac{1}{2}
In equation

\frac{1}{5} x-3 y=\frac{1}{6}

a2 = 1/5 ; b2 = – 3; c2 = –1/6

\frac{a_{1}}{a_{2}}= \frac{3}{5}\times \frac{5}{1}= 3;\frac{b_{1}}{b_{2}}= \frac{-1}{-3}= \frac{1}{3};\frac{c_{1}}{c_{2}}= \frac{-1}{-2}\times \frac{6}{1}= 3

For consistency either \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}or\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}} also here \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}

Hence given equations are consistent.  
Solution: (iii)
Equation are:   2ax + by = a
4ax + 2by = 2a
In equation
2ax + by – a = 0
a1 = 2a; b1 = b; c1 = –a
In equation
4ax + 2by – 2a = 0
a2 = 4a ; b2 = 2b; c2 = –2a

\frac{a_{1}}{a_{2}}= \frac{2a}{4a}= \frac{1}{2};\frac{b_{1}}{b_{2}}= \frac{b}{2b}= \frac{1}{2};\frac{c_{1}}{c_{2}}= \frac{-a}{-2a}= \frac{1}{2}
For consistency either \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}or \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} 

Also, here \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

Hence given equations are consistent
Solution: (iv)

Given, equations are   x + 3 y = 11
4x + 12y =22
In equation
x + 3y = 11
a1 = 1; b1 = 3; c1 = –11
In equation
4x + 12y = 22
a2 = 4 ; b2 = 12; c2 = –22
\frac{a_{1}}{a_{2}}= \frac{1}{4} ;\frac{b_{1}}{b_{2}}=\frac{3}{12}= \frac{1}{4};\frac{c_{1}}{c_{2}} = \frac{-11}{-22}= \frac{1}{2}
For consistency either \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}or \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} but here \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}

Hence given equations are not consistent.

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Do the following equations represent a pair of coincident lines? Justify your answer.

(i)3x+\frac{1}{7}y= 3; 7x + 3y = 7
(ii) –2x – 3y = 1 ; 6y + 4x = –2
(iii)  \frac{x}{2}+y+\frac{2}{5}= 0;4x+8y+\frac{5}{6}= 0

 

 

(i)Solution:
Equation are 3x + \frac{1}{7}y = 3
7x + 3y = 7
In equation
3x + \frac{1}{7}y – 3 = 0
a1 = 3 ; b1 = \frac{1}{7} ; c1 = –3
In equation
7x + 3y – 7 = 0
a2 = 7 ; b2 = 3; c2 = –7

\frac{a_{1}}{a_{2}}= \frac{3}{7};\frac{b_{1}}{b_{2}}= \frac{1}{21};\frac{c_{1}}{c_{2}}= \frac{-3}{-7}= \frac{3}{7}

For coincident lines \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}} but here \frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}

Hence the given pair of equations does not represent a pair of coincident lines. 

(ii)Solution:

Equation are   –2x – 3y = 1
4x + 6y = –2
In equation
–2x – 3y  – 1 = 0
a1 = –2 ; b1 = –3 ; c1 = –1
In equation
4x + 6y + 2 = 0
a2 = 4 ; b2 = 6; c2 = 2
\frac{a_{1}}{a_{2}}= \frac{-2}{4}= \frac{-1}{2};\frac{b_{1}}{b_{2}}= \frac{-3}{6}= \frac{-1}{2},\frac{c_{1}}{c_{2}}= \frac{-1}{2}

For coincident lines \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}} also here \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}
Hence the given pair of equation represents a pair of coincident lines.

Solution: (iii)
Equation are     \frac{x}{2}+y+\frac{2}{5}= 0           

4x + 8y + \frac{5}{16}= 0
In equation

\frac{x}{2}+y+\frac{2}{5}= 0
a1 = 1/2 ; b1 = 1 ; c1 = \frac{2}{5}
In equation
4x + 8y +  \frac{5}{16}=  0
a2 = 4 ; b2 = 8; c2 =  \frac{5}{16}

\frac{a_{1}}{a_{2}}= \frac{1}{8};\frac{b_{1}}{b_{2}}= \frac{1}{8};\frac{c_{1}}{c_{2}}\Rightarrow \frac{2}{5}\times \frac{16}{5}\Rightarrow \frac{32}{25}

For coincident lines \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}but here \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}

Hence the given pair of equations does not represent a pair of coincident lines.

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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+\frac{2}{3}y= 2

(i)Solution:
Equations are 2x + 4y = 3
12y + 6x = 6
In equation 2x + 4y – 3 = 0
a1 = 2, b1 = 4, c1 = –3
In equation 12y + 6x – 6 = 0
a2 = 6, b2 = 12, c2 = –6
\frac{a_{1}}{b_{2}}= \frac{2}{6}= \frac{1}{3},\frac{b_{1}}{b_{2}}= \frac{4}{12}= \frac{1}{3},\frac{c_{1}}{c_{2}}= \frac{-3}{-6}= \frac{1}{2}
For no solution \frac{a_{1}}{b_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}} also, Here we have \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}

Hence the given equation have no solution.
(ii)Solution:
Equations are x – 2y = 0
–2x + y = 0
In equation
x – 2y = 0
a1 = 1, b1 = –2, c1 = 0
In equation
–2x + y = 0
a2 = –2, b2 = 1, c2 = 0
\frac{a_{1}}{b_{2}}= \frac{1}{-2},\frac{b_{1}}{b_{2}}= \frac{-2}{1}= \frac{c_{1}}{c_{2}}= \frac{0}{0}
For no solution  but\frac{a_{1}}{b_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}} here

\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}
Hence the given pair of equations a unique solution.

(iii)Solution:
Equations are  3x + y = 3
2x+\frac{2}{3}y= 2
In equation
3x + y – 3 = 0
a1 = 3, b1 = 1, c1 = –3
In equation
2x+\frac{2}{3}y-2= 0
a2 = 2, b2= \frac{2}{3}, c2 = –2
\frac{a_{1}}{b_{2}}= \frac{3}{2},\frac{b_{1}}{b_{2}}= \frac{3}{2}= \frac{c_{1}}{c_{2}}= \frac{-3}{-2}= \frac{3}{2}

For no solution \frac{a_{1}}{b_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}} but here
\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}
Hence the given pair of equations have infinity many solutions.

 

 

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Write whether the following statements are True or False. Justify your answer. (-1, 7) is a point in the II quadrant.

Answer: True

Solution.

We know that signs in the II quadrant are (-, +) and here we have a point as (-1, 7)
i.e., the x-coordinate is negative and the y-coordinate is positive. 
Therefore the given statement is True.

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Write whether the following statements are True or False? Justify your answer.
A point lies on y-axis at a distance of 2 units from the x-axis. Its coordinates are (2, 0).

Answer: False

Solution.

We know that points that lie on the y-axis have coordinate in the form (0, y).
So we can say that x-coordinate should be zero.
The distance from x-axis will be equal to its y-coordinate.
So the point will be (0, 2) 
But here the point is given as (2, 0) so the statement is false. 

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Write whether the following statements are True or False? Justify your answer. The coordinates of a point whose ordinate is -1/2 and abscissa is 1 are -1/2 , 1.

Answer: False

Solution.


The abscissa is the x-axis (horizontal) coordinate
The ordinate is the y-axis (vertical) coordinate
Here ordinate is -1/2 and abscissa is 1.
So the coordinates are (1, -1/2) and not (-1/2, 1) .
Therefore the given statement is False.

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Write whether the following statements are True or False. Justify your answer.
Points (1, -1) and (-1. 1) lie in the same quadrant.

Answer: False

Solution.

False, point (1, -1) and (-1. 1) lie in the different quadrants. 
Point (1, -1) lies in the IV quadrant and point (-1. 1) lies in the II quadrant. 
Therefore the given statement is False.

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