Tag: pair of straight lines

Questions Related to pair of straight lines

If $P _{1},\ P _{2},\ P _{3}$ be the product of perpendiculars from $(0,0)$ to $xy+x+y+1=0$, $x^{2}-y^{2}+2x+1=0$, $2x^{2}+3xy-2y^{2}+3x+y+1=0$ respectively then?

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

  1. $P _{1} < P _{2}< P _{3}$

  2. $P _{3} < P _{2}< P _{1}$

  3. $P _{2} < P _{3}< P _{1}$

  4. $P _{1} < P _{3}< P _{2}$

Show answer
Correct Option: B
Explanation:
Given 
$xy+x+y+1=0$
comparing above eq with $ax^2+2hxy+by^2+2gx+2fy+c=0$
we get $h=\dfrac{1}{2},g=\dfrac{1}{2},f=\dfrac{1}{2},c=1$
Product of perpendicular from origin is $\left | \dfrac{c}{\sqrt{(a-b)^2+4h^2}} \right |$ 
$P _{1}=\left | \dfrac{1}{\sqrt{4\left ( \dfrac{1}{2} \right )^2}} \right |$
$P _{1}=\dfrac{1}{\sqrt{4\left ( \dfrac{1}{4} \right )}} $
$P _{1}=\dfrac{1}{\sqrt{1}} $
$P _{1}=1$
Now given eq $x^2-y^2+2x+1=0$ 
comparing above eq with $ax^2+2hxy+by^2+2gx+2fy+c=0$
we get $a=1,b=-1,h=0,g=1,f=0,c=1$
Product of perpendicular from origin is $\left | \dfrac{c}{\sqrt{(a-b)^2+4h^2}} \right |$ 
$P _{2}=\left | \dfrac{1}{\sqrt{\left ( 1+1 \right )^2}} \right |$
$P _{2}=\dfrac{1}{\sqrt{4}} $
$P _{2}=\dfrac{1}{2} $

Now given eq $2x^2+3xy-2y^2+3x+y+1=0$ 
comparing above eq with $ax^2+2hxy+by^2+2gx+2fy+c=0$
we get $a=2,b=-2,h=\dfrac{3}{2},g=\dfrac{3}{2},f=\dfrac{1}{2},c=1$
Product of perpendicular from origin is $\left | \dfrac{c}{\sqrt{(a-b)^2+4h^2}} \right |$ 
$P _{3}=\left | \dfrac{1}{\sqrt{\left ( 2+2 \right )^2-4\left ( \dfrac{3}{2} \right )^2}} \right |$
$P _{3}=\left | \dfrac{1}{\sqrt{16-4\left ( \dfrac{9}{4} \right )}} \right |$
$P _{3}=\left | \dfrac{1}{\sqrt{16-9}} \right |$
$P _{3}=\dfrac{1}{\sqrt{7}} $

SO $P _{3}< P _{2}< P _{1}$

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 $\displaystyle \frac{|c-c _{1}|}{\sqrt{(a^{2}+b^{2})}}$ 

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

  1. Assertion is true.

  2. Reason is true.

  3. Both Assertion and Reason are true.

  4. Neither Assertion nor Reason are true.

Show answer
Correct Option: B
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 $\Delta =\begin{vmatrix} 1 & \sqrt { 2 }  & 2\sqrt { 2 }  \ \sqrt { 2 }  & 4 & 1 \ 2\sqrt { 2 }  & 1 & 1 \end{vmatrix}=-2$

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

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

  1. $x^{2}+ax+6=0$

  2. $x^{2}+4ax+6=0$

  3. $x^{2}+4ax+2a^{2}+6=0$

  4. $x^{2}-4ax+6=0$

Show answer
Correct Option: C
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 $x^{2}+4ax+2a^{2}+6=0$

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

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

  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. $3x^{2} - 3y^{2} - 8xy - 10x - 15y - 20 = 0$

Show answer
Correct Option: B
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
$(x + 3y - 5)(3x - y - 5) = 0$
$\Rightarrow 3x^{2} - 3y^{2} + 8xy - 20x - 10y + 25 = 0$.

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

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

  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. $3x^{2}+8xy-3y^{2}-20x-10y+25=0$

Show answer
Correct Option: D
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
$3x^{2}+8xy-3y^{2}-20x-10y+25=0$

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

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

  1. Hyperbola

  2. A parabola

  3. An ellipse

  4. A ciircle

Show answer
Correct Option: A
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

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

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

  1. x = y = z

  2. y = 1, z = 1

  3. $\frac{x-2}{1}=\frac{y-2}{1}=\frac{z-2}{1}$

  4. $\frac{x-1}{1}=\frac{y-1}{1}=\frac{z-1}{1}$

Show answer
Correct Option: B
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 $\frac{x-1}{\frac{1}{2}}=\frac{y-1}{0}=\frac{z-1}{0}$
$\Rightarrow y=1,z=1$

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.

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

  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

Show answer
Correct Option: A
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.