# Tag: maths

### Questions Related to maths

The direction with $+x-axis$ in which a straight line will be drawn through the point $\left(1,2\right)$ so that its point of intersection with the line $x+y=4$ may be at a distance $\sqrt { \dfrac { 2 }{ 3 } }$ from the point $\left(1,2\right)$ can be:

1. ${15}^{o}$

2. ${30}^{o}$

3. ${45}^{o}$

4. ${75}^{o}$

Correct Option: A

The equation $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$ represents a pair of straight lines.

1. True

2. False

Correct Option: A

The curve satisfying the equation $\dfrac { dy }{ dx } =\dfrac { y(x+{ y }^{ 3 }) }{ x({ y }^{ 3 }-x) }$ and passing through the point (4, -2) is

1. ${ y }^{ 2 }=-2x$

2. ${ y }=-2x$

3. ${ y }^{ 3 }=-2x$

4. None of these

Correct Option: C

if the equation ${ 4x }^{ 2 }+2\sqrt { 3xy } +{ 2y }^{ 2 }-1=0$ becomes ${ 5x }^{ 2 }+{ y }^{ 2 }=1,\quad$  when the axes are rotar trough an angle ${ 45 }^{ 0 }$ , then the original  equation of the curve  is :'

1. ${ 15 }^{ 0 }$

2. ${ 30 }^{ 0 }$

3. ${ 45 }^{ 0 }$

4. ${ 60 }^{ 0 }$

Correct Option: A

The equation of a straight line passing through a point $(-5,4)$ and which cuts off an intercept of $\sqrt{2}$ units between the lines $x+y+1=0$ and $x+y-1=0$ is

1. $x-2y-13=0$

2. $2x-y+14=0$

3. $x-y+9=0$

4. $x-y+10=0$

Correct Option: C
Explanation:

Let the given point  be $A=(−5,4)$ and the given lines be $l _1 \rightarrow x+y+1=0$ and  $l _2\rightarrow x+y-1=0$

The  point $A$ lies on $l _1$

If segment $AM\perp l _2$ and $M$ lies on $l _2$, then, the distance $AM$ is given by,

$\Rightarrow AM=\dfrac{|−5+4−1|}{\sqrt{1^2+1^2}}=\dfrac{2}{\sqrt 2}=\sqrt 2$

$\Rightarrow$ This means that if $B$ is any point  on $l _2$ then  $AB>AM$. No line other than $AM$ cuts off an intercept of

length $\sqrt 2$ between $l _1$ and $l _2$.

$\Rightarrow$To determine the equation of $AM$, we need to find the co-ordinates of the Point $M$

Since, $AM\perp l _2$ and  the slope $l _2$ is $−1$, the slope of$AM$ must be $1$.  Also  $A(−5,4)$ lies on $AM$

By the point slope formula, the equation of the required line is

$\Rightarrow y−4=1(x−(−5))$

$\Rightarrow y-4=x+5$

$\Rightarrow x−y+9=0$

Equation of a straight line passing through the point $(4, 5)$ and equally inclined to the lines $3x=4y+7$ and $5y=12x+6$ is?

1. $9x-7y=1$

2. $9x+7y=71$

3. $7x+9y=73$

4. $7x-9y+17=0$

Correct Option: C

The number of values of $c$ such that the straight line $y=4x+c$ touches the curve $x^{2}+4y^{2}=4$, is

1. $2$

2. $0$

3. $1$

4. $\infty$

Correct Option: A

The equation of a straight line passing through the point (-5,4) and which cuts off an intercept of $\sqrt { 2 }$ unit between the lines $x+y+1=0$ and $x+y-1=0$ is:

1. $2x-y+14=0$

2. $3x+y+11=0$

3. $x-y+9=0$

4. $4x+5y=0$

Correct Option: C
Explanation:

Let the given point be $A=(-5,4)$ and the given lines be,

$L _1:x+y+1=0$ and
$L _2:x+y-1=0$
Observe that, $A\in L _1$.
If segment $AM\perp L _2,$ $M\in L _2,$ then, the distance $AM$ is given  by,

$\Rightarrow$  $AM=\dfrac{|-5+4-1|}{\sqrt{1^2+1^2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}$

This means that if $B$ is any point on $L _2,$ then, $AB>AM.$
In other words, no line other than $AM$ cuts off an intercept of length $\sqrt{2}$ between $L _1,$ and $L _2$ or $AM$ is the required line.
To determine the equation of $AM,$ we need to find the co-ordinates of the point $M$.
Since, $AM\perp L _2,$ and the slope of $L _2$ is $-1,$ the slope of $AM$ must be $1.$
Further, $A(-5,4)\in AM$
By the slope-point form the equation of the required line is,
$\Rightarrow$  $y-4=1(x-(-5))$
$\Rightarrow$  $y-4=x+5$
$\Rightarrow$  $x-y+9=0$

The equation of a straight line passing through the point (-5,  4) and which cuts off in intercept of $\sqrt { 2 }$ unit. between the lines $x+y+1=0$ and $x+y-1=0$ is:

1. $2x-y+14=0$

2. $3x+y+11=0$

3. $x-y+9=0$

4. $4x+5y=0$

Correct Option: C
Explanation:

Let the given point be $A\left( {{x} _{1}},{{y} _{1}} \right)=\left( -5,4 \right)$ and the given lines be

$x+y+1=0\,\,......\,\,\left( 1 \right)$

$x+y-1=0\,\,......\,\,\left( 2 \right)$

Observe that,

From equation (1) to,

Let AM is perpendicular distance

Then, $AM=\dfrac{\left| -5+4-1 \right|}{\sqrt{{{1}^{2}}+{{1}^{2}}}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}$

Let B is any point on equation $(2)$to,

From equation (2)

$x+y-1=0$

$y=-x+1$

On comparing that,

$y=mx+c$

Then, $m=-1$

Slope of perpendicular line is

${{m} _{1}}=\dfrac{-1}{m}=1$

Then, equation of line is

$y-{{y} _{1}}=m\left( x-{{x} _{1}} \right)$

$\Rightarrow y-4=1\left( x+5 \right)$

$\Rightarrow y-4=x+5$

$\Rightarrow x-y+5+4=0$

$\Rightarrow x-y+9=0$

A line passing through the points of intersection of $x+y=4$ and $x-y=2$ makes an angle $\tan^{-1}(3/4)$ with the x-axis. It intersects the parabola $y^2=4(x-3)$ at points $(x _1, y _1)$ and $(x _2, y _2)$ respectively. Then $|x _1-x _2|$ is equal to?
1. $\dfrac{16}{9}$
2. $\dfrac{32}{9}$
3. $\dfrac{40}{9}$
4. $\dfrac{80}{9}$