### Intersection of two straight lines - class-XI

 Description: intersection of two straight lines Number of Questions: 15 Created by: Rani Rajan Tags: coordinate geometry maths the straight line functions and graphs two dimensional analytical geometry
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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. $\dfrac{80}{9}$

Correct Option: A
$\displaystyle ax^{2}+2hxy+by^{2}=0$ represents a pair of straight lines through origin & angle between them is given by
$\displaystyle \tan \theta=\frac{2\sqrt{h^{2}-ab}}{a+b}$. If the lines are perpendicular then $\displaystyle a+b=0$ and the equation of bisectors is given by  $\displaystyle \frac{x^{2}-y^{2}}{a-b}=\frac{xy}{h}$
The general equation of second degree given by
$\displaystyle ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ represent a pair of straight lines if $\displaystyle \triangle =0$ or
$\displaystyle \begin{vmatrix}a&h &g \\ h&b &f \\ g&f &c \end{vmatrix}=0$ or $\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$
On the basis of above information answer the following question

If the lines joining origin to the points of intersection of the line $\displaystyle x +y = 1$ with the curve $\displaystyle x^{2}+y^{2}+x-2y-m = 0$ are perpendicular to each other, then value of $m$ is

1. $\displaystyle \frac{1}{2}$

2. $\displaystyle -\frac{1}{2}$

3. $\displaystyle 1$

4. $\displaystyle -1$

Correct Option: A
Explanation:

As the line $x+y=1$ intersect the curve,$\displaystyle x^{2}+y^{2}+x-2y-m=0$

$\displaystyle \Rightarrow x^{2}+y^{2}+\left ( x-2y \right )\left ( 1 \right )-m\left ( 1 \right )^{2}=0$

$\displaystyle \Rightarrow x^{2}+y^{2}+\left ( x-2y \right )\left ( x+y \right )-m\left ( 1 \right )^{2}=0$

$\displaystyle \Rightarrow x^{2}+y^{2}+\left ( x-2y \right )\left ( x+y \right )-m\left ( x+y \right )^{2}=0$

$\displaystyle \Rightarrow x^{2}+y^{2}+\left ( x^{2}-xy-2y^{2} \right )-m\left ( x^{2}+y^{2}+2xy \right )=0$

$\displaystyle \Rightarrow x^{2}\left ( 2-m \right )+y^{2}\left ( 1-2-m \right )+xy\left ( 1-2m \right )=0$      ......(*)

Now pair of straight lines given bye (*) will be $\perp$er if

$\displaystyle 2-m+1-2-m=0$ (Using $a+b=0$)

$\displaystyle \Rightarrow m=\frac{1}{2}$

If the lines joining the origin to the intersection of the line $y=mx+ 2$ and the curve $x^{2}+ y^{2}= 1$ are at right angles, then

1. $m^{2}=1$

2. $m^{2} = 3$

3. $m^{2}= 7$

4. $2m^{2} = 1$

Correct Option: C
Explanation:

Joint equation of the lines joining the origin and the point of intersection of the line $y =mx + 2$ and
the curve $x^{2} + y^{2}=1$ is
$\displaystyle x^{2} +y^{2} =\left( \frac{y-mx}{2}\right)^{2}$
$x^{2}(4 -m^{2} )+2mxy +3y^{2} = 0$
Since these lines are at right angles
$4 -m^{2} + 3 =0 \Rightarrow m^{2} = 7$

If the straight lines joining the origin and the points of intersection of the curve $5x^2 + 12xy -6y^2 + 4x -2y + 3 = 0$  and $x + ky -1 = 0$ are equally inclined to the co-ordinate axis, then the value of $k$

1. is equal to $1$

2. is equal to $-1$

3. is equal to $2$

4. does not exist in the set of real numbers

Correct Option: B
Explanation:

Homogenizing the curve with the help of the straight line.

$5x^2+12xy-6y^2+4x(x+ky) -2y(x+ky)+3(x+ky)^2 = 0$

$12x^2 + (10 + 4k + 6k) xy + (3k^2 -2k -6)y^2 = 0$

Lines are equally inclined to the coordinate axes

$\therefore$ coefficient of $xy = 0$

$\Rightarrow 10k + 10 = 0 \Rightarrow k = -1$

The lines joining the origin to the point of intersection of $3x^2 + mxy - 4x + 1 = 0$ and $2x + y - 1= 0$  are at right angles. Then which of the following is/are possible value/s of $m?$

1. $-4$

2. $4$

3. $7$

4. $3$

Correct Option: A,B,C,D
Explanation:

By application of the method of homogenization we get
$3x^2+mxy-4x(2x+y)+1(2x+y)^{2}$
$=3x^2+mxy-8x^2-4xy+4x^2+y^2+4xy$
$=-x^2+mxy+y^2$
$=0$
Hence the equations of the lines is given by $x^2-mxy-y^2=0$
Hence the lines are perpendicular for all values of $m.$

Find the equation of the lines joining the origin to the points of intersection of the curve $2x^2 + 3xy -4x + 1 = 0$ and the line $3x + y = 1$

1. $x^2-y^2-5xy=0$

2. $x^2+y^2-5xy=0$

3. $x^2+y^2+5xy=0$

4. $None\ of\ these$

Correct Option: A
Explanation:

Given equations of curve and the line are $2x^2 + 3xy -4x + 1 = 0$ and $3x + y = 1$.
Homogenising the curve with the line gives
$2x^2+3xy-4x(3x+y)+(3x+y)^2=0$
$\Rightarrow 2x^2+3xy-12x^2-4xy+9x^2+y^2+6xy=0$
$\Rightarrow y^2-x^2+5xy=0$
$\therefore$ The equation of the lines joining the origin to the points of intersection of the curve and hte line is $x^2-5xy- y^2=0$
Hence, option A.

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