Tag: condition for perpendicular and coincident lines and bisectors of angles

Questions Related to condition for perpendicular and coincident lines and bisectors of angles

The product of the perpendiculars from origin to the pair of lines $ a x ^ { 2 } + 2 h x y + b y ^ { 2 } + 2 g x + 2 f y + c = 0 $ is

business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

  1. $

    \frac { | c | } { \sqrt { ( a + b ) ^ { 2 } + 4 h ^ { 2 } } }

    $

  2. $

    \frac { | c | } { \sqrt { ( a + b ) ^ { 2 } - 4 h ^ { 2 } } }

    $

  3. $

    \frac { | c | } { \sqrt { ( a - b ) ^ { 2 } + 4 h ^ { 2 } } }

    $

  4. $

    \frac { | c | } { \sqrt { ( a - b ) ^ { 2 } - 4 h ^ { 2 } } }

    $

Show answer
Correct Option: A

If pair of lines $\displaystyle y^{2}+2hxy-9x^{2}=0$ and another pair of lines given by $\displaystyle ay^{2}+10xy+x^{2}=0$ have exactly one line common and other lines represented by them are perpendicular then

business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

  1. a = 9; h = -4

  2. a + h = 6

  3. angle between the lines represented by first pair is $\displaystyle \frac{\pi }{4}$

  4. all of these

Show answer
Correct Option: A
Explanation:

Let $ \displaystyle m _{1} $ and $ \displaystyle m _{2} $ be the slope of lines of first pair


$ \displaystyle \Rightarrow  m _{1}$ and  $ \displaystyle \Rightarrow  m _{2}$  be the slope of lines of second pair

Compairing pair of lines (y-$ \displaystyle \Rightarrow  m _{1} x $ )  (y-$ \displaystyle \Rightarrow  m _{2} x $ ) with $ \displaystyle y^{2}+2hxy-9x^{2}=0 $

and pair of lines (y-$ \displaystyle \Rightarrow  m _{1} x $ ) $ \displaystyle y+\frac{1}{m _{2}}x $ with $ \displaystyle ay^{2}+10xy+x^{2}=0 $ 

we get $ \displaystyle m _{1}=-1\, and\, m _{2}=9\Rightarrow a=9\,and\, h=-4 $

Two mutually perpendicular straight lines are drawn from the origin to form an isosceles triangle with the straight line $\displaystyle x\cos \alpha +y\sin \alpha -p=0$. Then the area of this triangle is

business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

  1. independent of $\displaystyle \alpha$

  2. independent of p

  3. independent of both $\displaystyle \alpha$ and p

  4. a function of $\displaystyle \alpha$ and p

Show answer
Correct Option: A
Explanation:

Two mutually perpendicular lines drawn from origin i.e X-axis and Y-axis 

$y=0$ and $x=0$
The area of triangle $=\dfrac{1}{2}\times x \times y$

Two of the lines represented by $x^{3}-6x^{2}y+3xy^{2}+dy^{3}=0$ are perpendicular for

business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

  1. all real values of $d$

  2. two real values of $d$

  3. three real values of $d$

  4. no real value of $d$

Show answer
Correct Option: B
Explanation:

Let $m _{ 1 },{ m } _{ 2 },{ m } _{ 3 }$ be the slopes of the three lines represnted by the given equation such that ${ m } _{ 1 }{ m } _{ 2 }=-1$


We have $\displaystyle m _{ 1 }{ m } _{ 2 }{ m } _{ 3 }=-\frac { 1 }{ d } $ so that 

$\displaystyle { m } _{ 3 }=\frac { 1 }{ d } $

Since $y={ m } _{ 3 }x\Rightarrow x=dy$ satisfies the given equation, we get

${ d }^{ 3 }-6{ d }^{ 2 }+3d+d=0\Rightarrow d\left( { d }^{ 2 }-6d+4 \right) =0$

If $d=0,$ the given equation represents the line $x=0$ and $x^2-6xy+3y^2=0$ which are not perpendicular

$\therefore d\neq 0$ and $\displaystyle d^2-6d+4=0\Rightarrow d=\frac { 6\pm \sqrt { 36-16 }  }{ 2 } =3\pm \sqrt { 5 } $

which gives two real values of $d$

The pair of lines represented by $3ax^{2}+5xy+(a^{2}-2)y^{2}=0$ are perpendicular to each other for 

business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

  1. two values of $a$

  2. for all values of $a$

  3. for one value of $a$

  4. for no value of $a$

Show answer
Correct Option: A
Explanation:

$3ax^2+5xy+(a^2-2)y^2=0$   are perpendicular to each other


$\therefore$ coefficient of $x^2+ $coefficient of $y^2=0$

$\Rightarrow 3a+a^2-2=0$

$\Rightarrow a^2+3a-2=0$


$\Rightarrow a=\dfrac{-3\pm\sqrt{9+8}}{2}$

so $a=\dfrac{-3+\sqrt{17}}{2}$ and $a=\dfrac{-3-\sqrt{17}}{2}$

Let $\Delta$ PQR be a right angled isoceles triangle which is right angled at $P(2,1)$. lf the equation of the line OR is $2x+y=3$, then the equation representing the pair of lines PQ and PR is 

business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

  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:

Since the triangle is isosceles, and right angled at P, therefore 
$PQ=PR$
And 
$PQ$ is perpendicular to $PR$.
The the combined equation of both PQ and PR should be of the form,
$ax^{2}-ay^{2}+bxy+cx+dy+e=0$
Since they are perpendicular and the coefficients of $x^{2}$ and $y^{2}$ will be equal and opposite.
Now the point P must satisfy the equation.
From the above options, the point P (2,1) only satisfies the equation in Option B.
Hence the correct Option is B.

The triangle ABC has medians AD, BE, CF. AD lies along the line $y = x + 3$ , BE lies along the line $y = 2x + 4$, AB has length $60$ and angle $C = 90$, then the area of ABC is

business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

  1. $400$

  2. $200$

  3. $100$

  4. $50$

Show answer
Correct Option: A
Explanation:
Given line 
$y=x+3-----(1)$
$y=2x+4-----(2)$
By shiftiing centroid to origin 
Thus now equation of median are $y=x$ and $y=2x$ 
Now the coordinates of A and B can be taken as $(a,a)$ and $B(b,2b)$
Using centroif formula $C(-a-b,-a-2b)$
Length of hypotenuse $AB=60$
$AB^2=(a-b)^2+(a-2b)^2$
$3600=a^2+b^2-2ab+a^2+4b^2-4ab$
$2a^2+5b^2-6ab=3600---(3)$
Slope of line AC from point $A,C$ is $m _{AC}=\dfrac{-a-2b-a}{-a-b-a}$
$m _{AC}=\dfrac{2a+2b}{2a+b}$
Slope of line BC from point $B,C$ is $m _{BC}=\dfrac{-a-2b-2b}{-a-b-b}$
$m _{BC}=\dfrac{a+3b}{a+2b}$
Angle $C=90$ means line AC and BC are perpendicular 
Hence $m _{AC}m _{BC}=-1$
$\dfrac{2a+2b}{2a+b}\times \dfrac{a+3b}{a+2b}=-1$
$2a^2+6ab+2ab+6b^2=-2a^2-4ab-ab-2b^2$
$4a^2+8b^2+13ab=0--------(4)$
Solving both $(3)$ and $(4)$ equations we get 
$ab=-\dfrac{800}{3}$
$Area of traingle=\dfrac{3}{2}(ab)=400$

The straight line joining the origin to the other two points of intersection of the curve whose equations are $\displaystyle ax^{2}+2hxy+2gx+by^{2}=0: : and: : a'x^{2}+2h'xy+b'y^{2}+2g'x=0$ will be at right angle if

business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

  1. $g(a' + b') - g'(a + b) = 0$

  2. $gg' = a'b' - ab$

  3. $g - g' = (a - b) (a' - b')$

  4. none of these

Show answer
Correct Option: A
Explanation:
Given eq of curves 
$ax^2+2hxy+2gx+by^2=0$----(1)
$a'x^2+2h'xy+2g'x+b'y^2=0$----(2)
$1=\dfrac{a'x^2+2h'xy+b'y^2}{-2g'x}$
Putting in eq (1)
$ax^2+2hxy+2gx\left ( \dfrac{a'x^2+2h'xy+b'y^2}{-2g'x} \right )+by^2=0$
$ax^2+2hxy-g\left ( \dfrac{a'x^2+2h'xy+b'y^2}{g'} \right )+by^2=0$
$ag'x^2+2hg'xy-ga'x^2-2gh'xy-gb'y^2+bg'y^2=0$
$x^2(ag'-a'g)+y^2(bg'-b'g)+2xy(hg'-gh')=0$
Here by perpendicularlly 
$ag'-a'g+bg'-b'g=0$
$(a+b)g'-g(a'+b')=0$
$g(a'+b')-g'(a+b)=0$

Which of the following pairs of straight lines intersect at right angles ?

business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

  1. $\displaystyle 2x^{2}=y\left ( x+2y \right ) $

  2. $\displaystyle \left ( x+y \right )^{2}=x\left ( y+3x \right )$

  3. $\displaystyle 2y\left ( x+y \right )=xy$

  4. $\displaystyle x=\pm 2y$

Show answer
Correct Option: A
Explanation:

For pair of straight lines, lines are perpendicular if $a+b=0$ (where $a$ is the coefficient of $x^2$ and $b$ is the coefficient of $y^2$ )

For option A,
$\displaystyle 2x^{2}=y\left ( x+2y \right ) $
$\Rightarrow 2x^{2}-xy-2y^{2}$
Here, $a+b=0$
Hence, the pair of straight lines intersect at right angles

For Option B,
$(x+y)^{2}=x(y+3x)$
$\Rightarrow -2x^{2}+xy+y^{2}=0$
$a+b \ne 0$ 

For Option C,
 $2y(x+y)=xy$
$\Rightarrow xy+2{y}^{2}=0$
$a+b \ne 0$ 

For Option D, $y=\pm 2x$
Clearly lines are not perpendicular.

Equation of line in the place $P=\equiv 2x-y+z-4=0$ which is perpendicular to the line I whose equation is $\dfrac{x-2}{1}=\dfrac{y-2}{-1}=\dfrac{z-3}{-2}$ and which passes through point of intersection of I and P is

business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

  1. $\dfrac{x-2}{3}=\dfrac{y-1}{5}=\dfrac{z-1}{-1}$

  2. $\dfrac{x-1}{3}=\dfrac{y-3}{5}=\dfrac{z-5}{-1}$

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

  4. $\dfrac{x-2}{2}=\dfrac{y-1}{-1}=\dfrac{z-1}{1}$

Show answer
Correct Option: A