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 the pair of lines $a{ x }^{ 2 }+2hxy+b{ y }^{ 2 }+2gx+2fy+c=0$ intersect on the y axis then

  1. $2fgh=b{ g }^{ 2 }+c{ h }^{ 2 }$
  2. $b{ g }^{ 2 }\neq c{ h }^{ 2 }$
  3. $abc=2fgh$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given pairs of lines

$S=ax^2+2hxy+by^2+2gx+2fy+c=0------(1)$ intersects at y-axis $x=0$ 

Let coordinate of point of intersection is $(0,y)$
To find the point of intersection: 
$\dfrac{d{S}}{d{x}}=0$

$2ax+2hy+2g=0$

Here point of intersection is $(0,y)$ SO above eq passes through $(0,y)$
$0+2hy=-2g$

$y=-\dfrac{g}{h}$
So Intersection point $\left (0,-\dfrac{g}{h} \right)$

Given pair passes through $\left (0,-\dfrac{g}{h} \right)$

$b\left( -\dfrac{g}{h}\right)^2+2f\left ( -\dfrac{g}{h} \right )+c=0$

$bg^2-2fgh+ch^2=0$

$2fgh=bg^2+ch^2$

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

If the equation of plane containing the line $\displaystyle \frac{-x-1}{3} = \frac{y-1}{2} = \frac{z+1}{-1}$ =1 and passing through the point (1, - 1, 0) is $ax+y+bz+c=0$, then (a+b+c) is equal to

  1. -3

  2. 3

  3. 0

  4. 2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The line is (x+1)/-3 = (y-1)/2 = (z+1)/-1. The plane passes through this line and the point (1, -1, 0). Find the normal vector by taking the cross product of the line's direction vector (-3, 2, -1) and the vector connecting a point on the line (-1, 1, -1) to (1, -1, 0).

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

Find the equation of the line passing through $(-3,5)$ and perpendicular to the line through the points $(2,5)$ and $(-3,6)$.

  1. $5x-2y+20=0$
  2. $x-5y+20=0$
  3. $5x-y+20=0$
  4. $5x+y+20=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The slope of the line, whose end points is $\left( {2,5} \right)$ and $\left( { - 3,6} \right)$ is,

$m = \frac{{6 - 5}}{{ - 3 - 2}}$

$ =  - \frac{1}{5}$

The two non-vertical lines are perpendicular to each other if they have the slopes as negative reciprocals of each other.

So, the slope of the line that is perpendicular to the line joining the points $\left( {2,5} \right)$ and $\left( { - 3,6} \right)$ is,

$ =  - \frac{1}{m}$

$ = 5$

The equation of line passing through the point $\left( { - 3,5} \right)$ with slope 5 is,

$\left( {y - 5} \right) = 5\left( {x - \left( { - 3} \right)} \right)$

$y - 5 = 5x + 15$

$5x - y + 20 = 0$

Therefore, the required equation of line is $5x - y + 20 = 0$.


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

If the distance between the pair of parallel lines ${x}^{2}+2xy+{y}^{2}-8ax-8ay-9{a}^{2}=0$ is $25\sqrt {2}$, then $a$ is

  1. $ \pm4$
  2. $\pm 2$
  3. $\pm 3$
  4. $\pm 5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents the general equation of pair of lines which are parallel to each other, then the distance between them is given by,


$d = \left|2\sqrt{\dfrac{g^2 - ac}{a(a+b)}} \right|$ (or)  $d = \left|2\sqrt{\dfrac{f^2 - ac}{b(a+b)}} \right|$

Here, the equation is, 
  $x^2 + 2xy + y^2 - 8ax - 8ay - 9a^2 = 0$


i.e., $a = 1, b = 1, c = -9a^2, h = 1, f = -4a, g = -4a $

$\because d = 25\sqrt{2}$
  

$\implies 25\sqrt{2} = \left|2\sqrt{\dfrac{(-4a)^2 - 1(-9a^2)}{1(1+1)}} \right|$

$\implies 25\sqrt{2} = \left|2\sqrt{\dfrac{16a^2 + 9a^2}{2}} \right|$

$\implies 25\sqrt{2} = \sqrt{2} (5a) $

$\therefore a = \pm 5$ (Ans)

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

If the pair of lines $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ intersecty on y-axis then

  1. $2fgh = b{g^2} + c{h^2}$
  2. $b{g^2} \ne c{h^2}$
  3. $abc = 2fgh$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Put $x = 0$ in the given equation
$\Rightarrow\,b{y}^{2}+2fy+c=0$.
For unique point of intersection ${f}^{2}−bc=0$
$\Rightarrow\,a{f}^{2}−abc=0$.
Since $abc+2fgh−a{f}^{2}−b{g}^{2}−c{h}^{2}=0$
$\Rightarrow\,2fgh−b{g}^{2}−c{h}^{2}=0$
$\therefore\,2fgh=b{g}^{2}+c{h}^{2}$
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The graph $y^2 + 2xy + 40 |x| = 400$ divides the plane into regions. Then the area of bounded region is

  1. $200$ sq. units
  2. $400$ sq. units
  3. $800$ sq. units
  4. $500$ sq. units
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For $x < 0$, the equation is
$(y - 20) (y+2x + 20) =0$
Hence, the area is $20 \times 40 = 800$ sq. units.
For $x\geq0$ the equation simplifies to  $y^2-400+2xy+40x=(y-20)(y+20)+2x(y+20)=(y+20)(y+2x-20)=0$
Thus, obtained lines make a quadrilateral $ABCD,$
Area of quadrilateral $ABCD =2\times$ Area of triangle $BCD $$=2\times\dfrac{1}{2}\times$ base $\times $height$ = 20\times 40=800$ sq.units

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

If $P\left( {1 + \frac{t}{{\sqrt 2 }},2 + \frac{t}{{\sqrt 2 }}} \right)$ be any point on a line then the range of value of $t$ for which the point $P$ lies between the parallel lines $x + 2y = 1$ and $2x + 4y = 15$ is

  1. $ - \dfrac{{4\sqrt 2 }}{5} < t < \dfrac{{5\sqrt 2 }}{6}$
  2. $ - \dfrac{{4\sqrt 2 }}{3} < t < \dfrac{{5\sqrt 2 }}{6}$
  3. $t < \dfrac{{ - 4\sqrt 2 }}{3}$
  4. $t < \dfrac{{5\sqrt 2 }}{6}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given:$P\left(1+\dfrac{t}{\sqrt{2}},2+\dfrac{t}{\sqrt{2}}\right)$ lies on $x+2y=1$
$\Rightarrow\,1+\dfrac{t}{\sqrt{2}}+2\left(2+\dfrac{t}{\sqrt{2}}\right)=1$
$\Rightarrow\,1+\dfrac{t}{\sqrt{2}}+4+\dfrac{2t}{\sqrt{2}}=1$
$\Rightarrow\,\dfrac{3t}{\sqrt{2}}=-4$
$\Rightarrow\,t=\dfrac{-4\sqrt{2}}{3}$
$P\left(1+\dfrac{t}{\sqrt{2}},2+\dfrac{t}{\sqrt{2}}\right)$ lies on $2x+4y=15$
$\Rightarrow\,2\left(1+\dfrac{t}{\sqrt{2}}\right)+4\left(2+\dfrac{t}{\sqrt{2}}\right)=15$
$\Rightarrow\,2+\dfrac{2t}{\sqrt{2}}+8+\dfrac{4t}{\sqrt{2}}=15$
$\Rightarrow\,\dfrac{6t}{\sqrt{2}}=5$
$\Rightarrow\,t=\dfrac{5\sqrt{2}}{6}$
Given:Range of value of $t$ for $t\in\left(\dfrac{-4\sqrt{2}}{3},\dfrac{5\sqrt{2}}{6}\right)$ lies between the parallel lines.
$\therefore\,\dfrac{-4\sqrt{2}}{3}<t<\dfrac{5\sqrt{2}}{6}$
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

The distance between the two lines represented by the equation $9x^2 - 24 xy + 16y^2 - 12 x + 16y - 12 = 0$ is

  1. $\dfrac{8}{5}$
  2. $\dfrac{6}{5}$
  3. $\dfrac{11}{5}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The above given equation
$9x^2-24xy+16y^2-12x+16y-12=0$ can be factorized as
$(3x-4y+2)(3x-4y-6)=0$
Since both the lines are parallel the distance between them will be
$d=\left|\dfrac{C _{2}-C _{1}}{\sqrt{3^2+4^2}}\right|$
$=\left|\dfrac{2-(-6)}{5}\right|$
$=\dfrac{8}{5}$

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

The equation $x^2y^2 - 9y^2- 6x^2 y + 54y = 0$ represents

  1. A pair of straight lines and a circle

  2. A pair of straight lines and a parabola

  3. A set of four straight lines forming a square

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x^2y^2-9y^2-6x^2y+54y=0$
$y^2(x^2-9)-6y(x^2-9)=0$
$(y^2-6y)(x^2-9)=0$
$y(y-6)(x+3)(x-3)=0$
Therefore the lines are
$y=0$
$y=6$
$x=3$
$x=-3$
The above set of lines represent a square of side $6$ units.

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

The product of the perepndiculars drawn from the point $\left(x _1,y _1\right)$ on the lines $ax^2+2hxy+by^2=0$ is

  1. $\displaystyle \frac { a{ { x } _{ 1 } }^{ 2 }+2h{ x } _{ 1 }{ y } _{ 1 }+b{ { y } _{ 1 } }^{ 2 } }{ \sqrt { { \left( a-b \right) }^{ 2 }+4{ h }^{ 2 } } } $
  2. $\displaystyle \frac { \left| a{ { x } _{ 1 } }^{ 2 }+2h{ x } _{ 1 }{ y } _{ 1 }+b{ { y } _{ 1 } }^{ 2 } \right| }{ \sqrt { { \left( a-b \right) }^{ 2 }+4{ h }^{ 2 } } } $
  3. $\displaystyle \frac { a{ { x } _{ 1 } }^{ 2 }-2h{ x } _{ 1 }{ y } _{ 1 }+b{ { y } _{ 1 } }^{ 2 } }{ \sqrt { { \left( a-b \right) }^{ 2 }+4{ h }^{ 2 } } } $
  4. $\displaystyle \frac { \left| a{ { x } _{ 1 } }^{ 2 }-2h{ x } _{ 1 }{ y } _{ 1 }+b{ { y } _{ 1 } }^{ 2 } \right| }{ \sqrt { { \left( a-b \right) }^{ 2 }+4{ h }^{ 2 } } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $\displaystyle y={ m } _{ 1 }x$ and $\displaystyle y={ m } _{ 2 }x$ be the two lines given by $\displaystyle{ x }^{ 2 }+2hxy+b{ y }^{ 2 }=0$ so that

$\displaystyle{ m } _{ 1 }+{ m } _{ 2 }=\frac { -2h }{ b } $ and $\displaystyle{ m } _{ 1 }{ m } _{ 2 }=\frac { a }{ b } $   ...(1)
The product of the perpendiculars drawn from $\displaystyle\left( { x } _{ 1, }{ y } _{ 1 } \right) $ on these lines

$\displaystyle=\frac { \left| { y } _{ 1 }-{ m } _{ 1 }{ x } _{ 1 } \right|  }{ \sqrt { 1+{ { m } _{ 1 } }^{ 2 } }  } .\frac { \left| { y } _{ 1 }-{ m } _{ 2 }{ x } _{ 1 } \right|  }{ \sqrt { 1+{ { m } _{ 2 } }^{ 2 } }  } $

$\displaystyle =\frac { { { y } _{ 1 } }^{ 2 }-\left( { m } _{ 1 }+{ m } _{ 2 } \right) { x } _{ 1 }{ y } _{ 1 }+{ m } _{ 1 }{ m } _{ 2 }{ { x }^{ 2 } } _{ 1 } }{ \sqrt { 1+{ { m }^{ 2 } } _{ 1 }+{ { m } _{ 2 } }^{ 2 }+{ { m } _{ 1 } }^{ 2 }{ { m } _{ 2 } }^{ 2 } }  } $

$=$$\displaystyle\dfrac { \left| { { { y } _{ 1 } }^{ 2 }-\left( { m } _{ 1 }+{ m } _{ 2 } \right) { x } _{ 1 }{ y } _{ 1 }+{ m } _{ 1 }{ m } _{ 2 }{ { { { x } _{ 1 } }^{ 2 } } } } \right|  }{ \sqrt { 1+{ \left( { m } _{ 1 }+{ m } _{ 2 } \right)  }^{ 2 }-2{ m } _{ 1 }{ m } _{ 2 }+{ { m } _{ 1 } }^{ 2 }{ { m } _{ 2 } }^{ 2 } }  } $

$\displaystyle=\dfrac { \left| { { y } _{ 1 } }^{ 2 }+\dfrac { 2h{ x } _{ 1 }y _1 }{ b } +\dfrac { a{ { x } _{ 1 } }^{ 2 } }{ b }  \right|  }{ \sqrt { 1+\dfrac { 4{ h }^{ 2 } }{ { b }^{ 2 } } -\dfrac { 2a }{ b } +\dfrac { { a }^{ 2 } }{ { b }^{ 2 } }  }  } $     (using(1) )

$\displaystyle=\frac { \left| { { a }x _{ 1 } }^{ 2 }+2h{ x } _{ 1 }{ y } _{ 1 }+{ { by } _{ 1 } }^{ 2 } \right|  }{ \sqrt { { \left( a-b \right)  }^{ 2 }+4{ h }^{ 2 } }  } $