Tag: distances and midpoints

Questions Related to distances and midpoints

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

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

  1. $2fgh=b{ g }^{ 2 }+c{ h }^{ 2 }$

  2. $b{ g }^{ 2 }\neq c{ h }^{ 2 }$

  3. $abc=2fgh$

  4. $None\ of\ these$

Show answer
Correct Option: A
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 $\left (0,-\dfrac{g}{h} \right)$

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

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

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

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

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

  1. -3

  2. 3

  3. 0

  4. 2

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Correct Option: B

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

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

  1. $5x-2y+20=0$

  2. $x-5y+20=0$

  3. $5x-y+20=0$

  4. $5x+y+20=0$

Show answer
Correct Option: C
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 $5x - y + 20 = 0$.


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

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

  1. $ \pm4$

  2. $\pm 2$

  3. $\pm 3$

  4. $\pm 5$

Show answer
Correct Option: D
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)

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

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

  1. $2fgh = b{g^2} + c{h^2}$

  2. $b{g^2} \ne c{h^2}$

  3. $abc = 2fgh$

  4. none of these

Show answer
Correct Option: A
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 $abc+2fgh−a{f}^{2}−b{g}^{2}−c{h}^{2}=0$
$\Rightarrow\,2fgh−b{g}^{2}−c{h}^{2}=0$
$\therefore\,2fgh=b{g}^{2}+c{h}^{2}$

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

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

  1. $200$ sq. units

  2. $400$ sq. units

  3. $800$ sq. units

  4. $500$ sq. units

Show answer
Correct Option: C
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

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

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

  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. $t < \dfrac{{5\sqrt 2 }}{6}$

Show answer
Correct Option: B
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.
$\therefore\,\dfrac{-4\sqrt{2}}{3}<t<\dfrac{5\sqrt{2}}{6}$

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

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

  1. $\dfrac{8}{5}$

  2. $\dfrac{6}{5}$

  3. $\dfrac{11}{5}$

  4. None of these

Show answer
Correct Option: A
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
$d=\left|\dfrac{C _{2}-C _{1}}{\sqrt{3^2+4^2}}\right|$
$=\left|\dfrac{2-(-6)}{5}\right|$
$=\dfrac{8}{5}$

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

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

  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

Show answer
Correct Option: C
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.

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

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

  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 } }  } $

Show answer
Correct Option: B
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 } }  } $