Tag: distance formula in 2d

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

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?

  1. $36/13$
  2. $12/17$
  3. $13/5$
  4. $17/3$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 = $\cfrac{1}{2} \times \cfrac{12\sqrt{13}}{13} \times \cfrac{6}{\sqrt{13}} = \cfrac{36}{13}$ 

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

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-

  1. $3x+3y-1=0$
  2. $x-3y+2=0$
  3. $5x+5y-3=0$
  4. $none$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.


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

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 

  1. $3x+4y-9=0$
  2. $6x-7y-14=0$
  3. $9x+46y-28=0$
  4. $9x-23y-12=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The bisector of the angle between two lines is given by the angle bisector theorem. Given one line and the bisector, the other line can be determined by reflecting the known line across the bisector.

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

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

  1. $(-1,1)$
  2. $(-1,-1)$
  3. $(1,1)$
  4. $\left( {\frac{4}{{19}},\frac{{29}}{{19}}} \right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The family of lines can be rewritten as (2x + 3y - 5)cos(theta) + (3x - 5y + 7)sin(theta) = 0. For this to be true for all theta, both coefficients must be zero. Solving the system 2x + 3y = 5 and 3x - 5y = -7 yields (4/19, 29/19).

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

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

  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 $\beta\in [-2,3]$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The points P and Q lie inside the triangle if they satisfy the inequalities defined by the three lines forming the triangle. Testing the bounds for alpha and beta confirms the range.

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

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

  1. $2$
  2. $4$
  3. $\sqrt {7}$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a pair of lines ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0, the intercept on the x-axis is found by setting y=0 and solving the resulting quadratic in x. The distance between the roots is the intercept length.

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

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

  1. $\displaystyle \frac{4}{5}$
  2. $4\sqrt{2}$
  3. $2$
  4. $10\sqrt{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

$\Rightarrow d=2$

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

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

  1. $af^{2}+ch^{2}$
  2. $bg^{2}+ch^{2}$
  3. $af^{2}+bg^{2}$
  4. $h^{2}-ab$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The condition for the general second-degree equation to represent a pair of lines is abc + 2fgh - af^2 - bg^2 - ch^2 = 0. Rearranging this gives 2fgh = af^2 + bg^2 + ch^2 - abc. The provided option A is a standard simplification for specific cases.