Tag: distances and midpoints

Questions Related to distances and midpoints

If a pair of perpendicular straight lines drawn through the origin forms an isosceles triangle with the line $2x+3y=6$, then area of the triangle so formed is?

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $36/13$

  2. $12/17$

  3. $13/5$

  4. $17/3$

Show answer
Correct Option: A
Explanation:

As the lines are perpendicular and form an isosceles triangle the other two angles must be $45^\circ$

Let the slope of the line be m
$tan 45^\circ = \Bigg|\cfrac{-\cfrac{2}{3}-m}{1-\cfrac{2m}{3}}\Bigg| = 1$
$m = -5$ and the other line slope =$\cfrac{1}{5}$
Lines 
$y +5x= 0$ and $5y =x$
Intersection points $(0,0)$ , $(-\cfrac{6}{13} , \cfrac{30}{13})$ and $(\cfrac{30}{13} , \cfrac{6}{13})$
Perpendicular distance from origin to line $2x+3y=6$ is $\cfrac{|0+0-6|}{\sqrt{2^2+3^2}} = \cfrac{6}{\sqrt{13}}$
Distance between the points are $\sqrt{\Bigg(\cfrac{36}{13}\Bigg)^2+\Bigg(\cfrac{24}{13}\Bigg)^2} = \cfrac{\sqrt{1872}}{13} = \cfrac{12\sqrt{13}}{13}$
Area = $\cfrac{1}{2} \times \cfrac{12\sqrt{13}}{13} \times \cfrac{6}{\sqrt{13}} = \cfrac{36}{13}$ 

The line $x+3y-2=0$ bisects the angle between a pair of straight lines of which one has equation $x-7y+5=0$. The equation of the other line is-

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $3x+3y-1=0$

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

  3. $5x+5y-3=0$

  4. $none$

Show answer
Correct Option: C
Explanation:
we have
$L _1 =x+3y-2=0---(1)$

$L _2=x-7y+5=0---(2)$

now,

we know that

family of line through the given lines is

$L=L _2+\lambda L _2=0$

$=x-7y+5+\lambda (x+3y-2)=0---(3)$

Distance of any point ray $(2,0)$ on the line $x+3y-2=0$ from the lime 
$x-7y+5=0$ and the line $L=0$ must be same
so,

$\Rightarrow \ \left |\dfrac {2+5}{\sqrt {50}}\right | = \left |\dfrac {2+2\lambda +5-2\lambda}{\sqrt {(1+\lambda)^2+(3\lambda -7)^2}}\right|$

$\Rightarrow \ \dfrac {7}{\sqrt {50}}=\dfrac {7}{\sqrt {(1+\lambda)^2 +(3\lambda -7)^2}}$

$\Rightarrow \ 10\lambda^2-40\lambda =0$

$10\lambda (\lambda -4)=0$

$\lambda =0,\ \lambda =4$

then, put $\lambda =4$ in equation $(3)$ and we get

$L=x-7y+5+4(x+3y-2)=0$

$L=x-7y+5+4x+12y-8=0$

$L=5x+5y-3=0$

Hence this is the answer.


One of the lines of $-3x^{2}+2xy+y^{2}=0$ is parallel to $lx+y+1=0$ then $l=$

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $0$

  2. $1$

  3. $2$

  4. $-1$

Show answer
Correct Option: D
Explanation:

${ y }^{ 2 }+3x-xy-3{ x }^{ 2 }=0\ y(y+3x)-x(y+3x)=0\ (y+3x)(y-x)=0$

lines are $x=y$ and $-3x$
On comparing , we get 
$l=3$ or $l=-1$

If the straight line $2x+3y+1=0$ bisects the  angle between a pair of lines ,one of which in this pair is $3x+2y+4=0$, then the equation of the other line in that pair of line is 

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $3x+4y-9=0$

  2. $6x-7y-14=0$

  3. $9x+46y-28=0$

  4. $9x-23y-12=0$

Show answer
Correct Option: C

If $\theta $ is the parameter,then the family of lines respectedby $\left( {2\cos \theta  + 3\sin \theta } \right)x + \left( {3\cos \theta  - 5\sin \theta } \right)y - \left( {5\cos \theta  - 7\sin \theta } \right) = 0$: are concurrent at the point

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $(-1,1)$

  2. $(-1,-1)$

  3. $(1,1)$

  4. $\left( {\frac{4}{{19}},\frac{{29}}{{19}}} \right)$

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Correct Option: D

A triangle ${ABC}$ is formed by the lines $2x-3y-6=0$; $3x-y+3=0$ and $3x+4y-12=0$. If the points $P(\alpha,0)$ and $Q(0,\beta)$ always lie on or inside the $\triangle {ABC}$, then

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $\alpha \in [-1,2]$ and $\beta\in [-2,3]$

  2. $\alpha \in [-1,3]$ and $\beta\in [-2,4]$

  3. $\alpha \in [-2,4]$ and $\beta\in [-3,4]$

  4. $\alpha \in [-1,3]$ and $\beta\in [-2,3]$

Show answer
Correct Option: D

The distance the lines 3x +4 y = 9 and 6x +8y = 15 is = 

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. 3/2

  2. 3/10

  3. 6

  4. None of these

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Correct Option: D

In the equation $2x^{2}+2hxy+6y^{2}-4x+5y-6=0$ represent a pair of straight lines then the length of intercept on the $x-$axis cut by the lines is

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $2$

  2. $4$

  3. $\sqrt {7}$

  4. $0$

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Correct Option: B

The distance between the lines given by $(x+7y)^{2}+4 \sqrt{2}(x+7y)-42=0,$ is

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $\displaystyle \frac{4}{5}$

  2. $4\sqrt{2}$

  3. $2$

  4. $10\sqrt{2}$

Show answer
Correct Option: C
Explanation:

Given equation of pair of straight lines is 

$(x+7y)^{2}+4 \sqrt{2}(x+7y)-42=0$

$\Rightarrow x^{2}+49y^{2}+14xy+4\sqrt{2}x+28\sqrt{2}y-42=0$

Here $a=1, b=49, h=7, g=2\sqrt{2},f=14\sqrt{2},c=-42$
Here $h^{2}=ab$

The given equation represents a pair of parallel lines.

$\displaystyle d=2\sqrt{\frac{g^{2}-ac}{a(a+b)}}$

$\Rightarrow d=2$

If the pair of lines $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ intercept on the $x-$axis, then $2fgh=$

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $af^{2}+ch^{2}$

  2. $bg^{2}+ch^{2}$

  3. $af^{2}+bg^{2}$

  4. $h^{2}-ab$

Show answer
Correct Option: A