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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 x axis but it is parallel to y-axis.Hence the given statement is false.

#### 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 $\lambda$x + 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.

#### 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.

#### 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.

Solution.

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

#### 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).

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.

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.

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

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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