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

Assertion (A): The distance between the lines represented by $x^{2}+2\sqrt{2}xy+2y^{2}+4\sqrt{2}x+4y+1=0$ is 2 
Reason (R): Distance between the lines $ax+by+c=0$ and $ax+by+c _{1}=0$ is $\displaystyle \frac{|c-c _{1}|}{\sqrt{(a^{2}+b^{2})}}$ 

  1. Assertion is true.

  2. Reason is true.

  3. Both Assertion and Reason are true.

  4. Neither Assertion nor Reason are true.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Reason is true 
Assertion
$x^{ 2 }+2\sqrt { 2 } xy+2y^{ 2 }+4\sqrt { 2 } x+4y+1=0$ represents a pair of lines when $\Delta =0$
But $\Delta =\begin{vmatrix} 1 & \sqrt { 2 }  & 2\sqrt { 2 }  \ \sqrt { 2 }  & 4 & 1 \ 2\sqrt { 2 }  & 1 & 1 \end{vmatrix}=-2$

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

lf the expression $3x^{2}+2pxy+2y^{2}+2ax-4y+1$ can be resolved into two linear factors, then $p$ must be a root of the equation

  1. $x^{2}+ax+6=0$
  2. $x^{2}+4ax+6=0$
  3. $x^{2}+4ax+2a^{2}+6=0$
  4. $x^{2}-4ax+6=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If $3x^{2}+2pxy+2y^{2}+2ax-4y+1=0.$, then.
$\Delta =abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$
$=3(2)(1)+2(-2)(a)(p)-3(-2)^{2}-2(a)^{2}-1(p)^{2}=0$
$6-4ap-12-2a^{2}-p^{2}=0$
$p^{2}+4ap+2a^{2}+6=0$
$\Rightarrow p $ is a solution of $x^{2}+4ax+2a^{2}+6=0$

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

Let $PQR$ be a right angled isosceles triangle, right angled at $P(2, 1)$. If the equation of the line $QR$ is $2x + y = 3$. Then the equation representing the pair of lines $PQ$ and $PR$ is

  1. $3x^{2} - 3y^{2} + 8xy + 20x + 10y + 25 = 0$
  2. $3x^{2} - 3y^{2} + 8xy - 20x - 10y + 25 = 0$
  3. $3x^{2} - 3y^{2} + 8xy + 10x + 15y + 20 = 0$
  4. $3x^{2} - 3y^{2} - 8xy - 10x - 15y - 20 = 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equations of $PQ$ and $PR$ are given by
$y - 1 = \dfrac {-2\mp \tan 45^{\circ}}{1\pm (-2)\tan 45^{\circ}} (x - 2)$


$\Rightarrow y - 1 = \left (\dfrac {-2\mp 1}{1\pm 2}\right ) (x - 2)$

$\Rightarrow y - 1 = -\dfrac {1}{3} (x - 2)$ and $y - 1 = 3(x - 2)$

$\Rightarrow x + 3y = 5$ and $3x - y = 5$
The combined equation of these two lines is
$(x + 3y - 5)(3x - y - 5) = 0$
$\Rightarrow 3x^{2} - 3y^{2} + 8xy - 20x - 10y + 25 = 0$.

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

Let $\triangle PQR$ be a right angled isosceles triangle, right angled at $P(2, 1)$. If the equation of the side $QR$ is $2x + y = 3$, then the combined equation of sides $PQ$ and $PR$ is

  1. $3x^{2}-8xy+3y^{2}+20x-10y-25=0$
  2. $3x^{2}+8xy-3y^{2}+20x+10y+25=0$
  3. $3x^{2}-8xy+3y^{2}-20x-10y+25=0$
  4. $3x^{2}+8xy-3y^{2}-20x-10y+25=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Slopes of the line $PQ$ and $PR$ are
$\tan \left(\theta +\dfrac\pi4\right)=\dfrac{1+\tan \theta }{1-\tan \theta }=\dfrac{1-2}{1+2}=-\dfrac{1}{3}$ and $3$
$\therefore $ Equations of $PQ$ and $PR$ are $3y + x - 5 = 0$ and $y-  3x + 5 = 0$
$\therefore $ Combined equation of $PQ$ and $PR$ is
$3x^{2}+8xy-3y^{2}-20x-10y+25=0$

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

The locus of a point which moves such that the square of its distance from the base of an isosceles triangle is equal to the rectangle under its distances from the other two sides is

  1. Hyperbola

  2. A parabola

  3. An ellipse

  4. A ciircle
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
If the triangle is PQR, with $PQ = PR$, take P, Q, R to be the points $(0,b), (–a,0), (a,0)$ respectively.
The equation of the line PQ is 
$y-b=\dfrac{b}{+a}(x)$
$ay-ab=bx$
$bx-ay+ab=0$
and the equation of PR is
$y-b=\dfrac{b}{-a}(x)$
$-ay+ab=bx$
$bx+ay-ab=0$
and the equation of QR is
$y=\dfrac{0}{2a}(x)$
$y=0$
To find the locus of a point $X(h,k)$ which moves so that the square of its distance from QR is equal to the product of its distances from $PQ$ and $PR$. The distance from $X(h,k)$ to PQ is 

$d _{1}=\left | \dfrac{bh-ak+ab}{\sqrt{a^2+b^2}} \right |$
and the distance from $X(h,k)$ to PR is
$d _{2}=\left | \dfrac{bh+ak-ab}{\sqrt{a^2+b^2}} \right |$
The distance from $X(h,k)$ to QR is $|k|$
So according to question 
The distance of $X$ to QR=product of distances from $X$ to PQ and PR 
$k=d _{1}d _{2}$

$k=\dfrac{bh-(ak-ab)}{\sqrt{a^2+b^2}}\times\dfrac{bh+(ak-ab)}{\sqrt{a^2+b^2}}$

$k=\dfrac{b^2h^2-(a^2k^2+a^2b^2-2a^2kb)}{a^2+b^2}$

$a^2k+b^2k=b^2h^2-a^2k^2-a^2b^2+2a^2bk$
Putting $h=x,k=y$
$b^2x^2+(2a^2+b^2)y^2+2a^2by-a^2b^2=0$
Hence above equation represents the pair of straight lines

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

If G is the centroid and O is the circumcentre of the triangle with vertices (1, 2, 0), (0, 0, 2) and (2, 1, 1), then equation/s of line OG is/are

  1. x = y = z

  2. y = 1, z = 1
  3. $\frac{x-2}{1}=\frac{y-2}{1}=\frac{z-2}{1}$
  4. $\frac{x-1}{1}=\frac{y-1}{1}=\frac{z-1}{1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A(1, 2, 0), B (0, 0, 2) and C(2, 1, 1)
$\therefore$ G(1, 1, 1)
$AB^{2} = 1 + 4 + 4 = 9$, $AC^{2} = 1 + 1 + 1 = 3$ and $BC^{2} = 4 + 1 + 1 = 6$
$\therefore$ $AB^{2} =AC^{2} + BC^{2}$
$\therefore$ $\Delta $ ABC is right angled at C
$\therefore$ O is the mid point of AB
$\therefore$ coordinates of O are $\left ( \frac{1}{2},1,1 \right )$
$\therefore$ equation of OG are $\frac{x-1}{\frac{1}{2}}=\frac{y-1}{0}=\frac{z-1}{0}$
$\Rightarrow y=1,z=1$

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

STATEMENT-1  :There lies exactly $3$ unique points on the curve $8{ x }^{ 3 }+{ y }^{ 3 }+6xy=1$ which  form an equilateral triangle.

STATEMENT-2  :  The curve $8{ x }^{ 3 }+{ y }^{ 3 }+6xy=1$ consists  of  a  straight  line and a point which does not lie on the line.

  1. STATEMENT -1 is True, STATEMENT -2 is True; STATEMENT-2 is a correct explanation for STATEMENT - 1

  2. STATEMENT -1 is True,STATEMENT -2 is True; STATEMENT -2 is NOT a correct explanation for STATEMENT - 1

  3. STATEMENT-1 is True, STATEMENT-2 is False

  4. STATEMENT-1 is False, STATEMENT-2 is True
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have, 
$y^3+8x^3 = 1 -6xy $

adding both the sides  $ 6xy^2+12x^2y$,
$y^3 + 6xy^2 + 12 x^2y + 8x^3 = 1 - 6xy +6xy^2 +12x^2y$
$(y+2x)^3 = 1 - 6xy +6xy^2 +12x^2y$
$(y+2x)^3 -1^3 = 6xy(-1+y+2x)$
$(y+2x-1)((y+2x)^2 + (y+2x) +1 ) = 6xy(2x+y-1)$
$(y+2x-1)(y^2+4x^2+4xy + y +2x +1 ) = 6xy(2x+y-1)$
$(y+2x-1)(y^2+4x^2 -2xy +y+2x+1) = 0$
$(y^2+4x^2 -2xy +y+2x+1)=0$
${ y }^{ 2 }+y(1-2x)+4{ x }^{ 2 }+2x+1=0$
$D={ (1-2x) }^{ 2 }-4(4{ x }^{ 2 }+2x+1)=-3{ (2x+1) }^{ 2 }$
For real $y$
$D=0$
and $x=-\dfrac12$ and $y=1$.
So there is only one point which doesn't lie on the straight line.
Hence Assertion and Reason both are correct and reason is correct explanation.