Equations of lines parallel to the x-axis and y-axis - class-IX

 Description: equations of lines parallel to the x-axis and y-axis Number of Questions: 15 Created by: Niharika Sharma Tags: linear equations in two variables lines maths fundamentals
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If $2^{2x-y}=32$ and $2^{x+y}=16$ then $x^2+y^2$ is equal to

1. 9

2. 10

3. 11

4. 13

Correct Option: B
Explanation:

Given,

${ 2 }^{ 2x-y }=32$
${ 2 }^{ x+y }=16$
these equations can be writtn as
${ 2 }^{ \left( 2x-y \right) }={ 2 }^{ 5 }$
$\Rightarrow 2x-y=5$.......................eq1
${ 2 }^{ x+y }={ 2 }^{ 4 }$
$\Rightarrow x+y=4$...........................eq2
on adding both equations
$x+y=4$
$2x-y=5$
_______________________
$3x=9$
$x=3$
putting the value of x we get the value of y
$3+y=4$
$y=1$
putting the values of x and y we get
${ x }^{ 2 }+{ y }^{ 2 }$
${ 3 }^{ 2 }+{ 1 }^{ 2 }=9+1=10$

If the expression $\displaystyle (x+y)^{-1}. (x^{-1}+y^{-1})(xy^{-1}+x^{-1}y)^{-1}$ is simplefied it takes the form of which one of the following ?

1. x+y

2. $\displaystyle( x^{2}+y^{2})^{-1}$

3. xy

4. $\displaystyle x^{2}+y^{2}$

Correct Option: B
Explanation:
$\left ( x+y \right )^{-1}\left ( \frac{1}{x}+\frac{1}{y} \right )^{-1}\left ( xy^{-1}+x^{-1}y \right )^{-1}$
$\frac{1}{x+y}\left ( \frac{1}{x}+\frac{1}{y} \right )\left ( \frac{x}{y}+\frac{y}{x}^{-1} \right )$
=$\frac{1}{x+y}\left ( \frac{x+y}{xy} \right )\left ( \frac{x^{2}+y^{2}}{xy} \right )^{-1}$
=$\frac{1}{x+y}\times \frac{x+y}{xy}\times \frac{xy}{x^{2}+y^{2}}$
=$\left ( \frac{1}{x^{2}+y^{2}} \right )=(x^{2}+y^{2})^{-1}$

Any point on $x$ axis is of the form

1. $(x,y)$

2. $(0,y)$

3. $(0,0)$

4. $(x,0)$

Correct Option: D
Explanation:

Any point on $x$ axis is of the form $(x,0)$ and any point on $y$ axis is of the form $(0,y)$.

For example, $(5,0)$ is a point that lies on $x$ axis.

Any point on $y$ axis is of the form

1. $(x,y)$

2. $(0,y)$

3. $(y,y)$

4. $(x,0)$

Correct Option: B
Explanation:

Any point on $x$ axis is of the form $(x,0)$ and any point on $y$ axis is of the form $(0,y)$.

For example, $(0,5)$ is a point that lies on $y$ axis.

$y = 2$ is a line

1. Parallel to $x$-axis

2. Parallel to $y$-axis

3. Passing through origin

4. None

Correct Option: A
Explanation:

In general, $x=a$ is a line lies on $x$ axis and parallel to $y$ axis and $y=b$ is a line lies on $y$ axis and parallel to $x$ axis.

Hence, $y=2$ is a line parallel to $x$ axis.

The line $x - 7 = 0$ is

1. Parallel to $y$-axis.

2. Parallel to $x$-axis.

3. Passing through the origin.

4. None of these.

Correct Option: A
Explanation:

The line will have a constant value of $x=7$,i.e, only $y$ will vary so it is parallel to $y-axis$.

The angles between the lines $3 x + y - 7 = 0 \text { and } x + 2 y + 9 = 0$ is:

1. $\dfrac { \pi } { 3 }$

2. $\dfrac { \pi } { 6 }$

3. $\dfrac { \pi } { 2 }$

4. $\dfrac { \pi } { 4 }$

Correct Option: D
Explanation:
Slope of the line $y=-3x+7$ is $-3$
Slope of the line $2y=-x-9$ or $y=\dfrac{-1}{2}x-\dfrac{9}{2}$ is $-\dfrac{1}{2}$

$\therefore\,{m} _{1}=-3,\,{m} _{2}=-\dfrac{1}{2}$

Angle between the line $=\tan{\theta}=\left|\dfrac{{m} _{1}-{m} _{2}}{1+{m} _{1}{m} _{2}}\right|$

$=\left|\dfrac{-3+\dfrac{1}{2}}{1+\left(-3\right)\left(-\dfrac{1}{2}\right)}\right|$

$=\left|\dfrac{\dfrac{-5}{2}}{1+\dfrac{3}{2}}\right|$

$=\left|\dfrac{\dfrac{-5}{2}}{\dfrac{2+3}{2}}\right|$

$=\left|\dfrac{\dfrac{-5}{2}}{\dfrac{5}{2}}\right|$

$\tan{\theta}=1$

$\therefore\,\theta=\dfrac{\pi}{4}$

Draw a graph for the following equations:

1. y = -x

2. y = -2.5x

3. Both A and B

4. None of these

Correct Option: A

If the line segment joining $(2,3)$ and $(-1,2)$ is divided in the ratio $3:4$ by the graph of the equation $x+2y=k$, the value of $k$ is

1. $\dfrac{5}{7}$

2. $\dfrac{31}{7}$

3. $\dfrac{36}{7}$

4. $\dfrac{41}{7}$

Correct Option: A

If $4x+3y=120$, find how many non-negative integer solutions are possible?

1. $1$

2. $11$

3. Infinite

4. None of these

Correct Option: D
Explanation:

We can write the equation in another form $4(x-3)+3(y+4)=120$
Now, we make a table

x 30 27 24 21 18 15 12 9 6 3 0
y 0 4 8 12 16 20 24 28 32 36 40

We observe a patter that $x$ reduces by $3$ and $y$ increases by $4$. But both $x$ and $y$ cannot be negative or $0$. So, the value of $x=0$ and $y=0$ is ruled out. Hence, nine such positive integer solutions are possible.
Alternately, we can write the given equation as $4(x+3)+3(y-4)=120$. We get the same values of $x$ and $y$

In the equation $2x - 3y = 12$. The value of $x$ and $y$ that satisfy the equation are-

1. $(5, 3)$

2. $(2, 3)$

3. $(6, 0)$

4. $(3,12)$

Correct Option: C
Explanation:

Put $\displaystyle x=6$ and $\displaystyle y=0$ in given equation.
$\displaystyle 2x-3y=2\times 6-3\times 0$
=12-0
=12
Hence, the value of x and y that satisfy the equation are (6,0).

The graph of $x =8$ represents:

1. line parallel to $y$-axis and at a distance of $8$ units.

2. line parallel to $x$-axis and at a distance of $8$ units.

3. line parallel to $y$-axis and at a distance of $0$ units.

4. None of these

Correct Option: A
Explanation:

The line $x=8$ or a point $(8,0)$ lies on the $x$-axis whwich is parallel to $y$-axis and at a distance of $8$ units.

$(5,0)$ is a point that lies on

1. $y$-axis

2. $x$-axis

3. $y=x$

4. $y=5x$

Correct Option: B
Explanation:

Any point on $x$ axis is of the form $(x,0)$ and any point on $y$ axis is of the form $(0,y)$.

Therefore, $(5,0)$ is a point that lies on $x$ axis.

The graph of $|x| + |y|$ consists of

1. one straight line

2. a pair of straight lines

3. the sides of a square

4. a circle

5. a point

Correct Option: C
Explanation:

Given $|x|+|y|$

1. For $x>0$ and $y>0$ , we have $x+y$
2. For $x>0$ and $y<0$ , we have $x-y$
3. For $x<0$ and $y>0$ , we have $-x+y$
4. For $x<0$ and $y<0$ , we have $-x-y$
5. observe that the four equations will form a square

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