Mathematics

Straight Lines and Angles

162 Questions

Straight lines and angles are core components of coordinate geometry. This topic evaluates angle measures between intersecting lines, direction ratios, and perpendicular distances. Mastery of these mathematical concepts is necessary for high scores in quantitative exams.

Angle between linesDirection ratiosAngle bisectorsPerpendicular distanceSlope differences

Straight Lines and Angles Questions

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

If one of the lines of is $my^{2}+\left ( 1-m^{2} \right )xy-mx^{2}=0$ is a bisector of the angle between the lines $\displaystyle xy = 0,$ then $m$ is

  1. $\displaystyle1$
  2. $\displaystyle2$
  3. $\displaystyle-\frac{1}{2}$
  4. $\displaystyle-1$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

Given pair of lines is
$\displaystyle { my }^{ 2 }+\left( 1-{ m }^{ 2 } \right) xy-{ mx }^{ 2 }=0\Rightarrow m{ \left( \frac { y }{ x }  \right)  }^{ 2 }+{ \left( 1-m \right)  }^{ 2 }\frac { y }{ x }- m=0$    ...(1)

Lines $xy=0$ are $x=0$ and $y=0.$
i.e the axes bisector of angle between the axes are $y=x$ and $y=-x$ $\displaystyle \Rightarrow \frac { y }{ x } =1$ or $\displaystyle \frac { y }{ x } =-1$

If $\displaystyle \frac { y }{ x } =1$ is represented by (1), then
$m{ \left( 1 \right)  }^{ 2 }+\left( 1-{ m }^{ 2 } \right) \left( -1 \right) -m=0\Rightarrow 1-{ m }^{ 2 }=0\Rightarrow m=\pm 1$

If $\displaystyle \frac { y }{ x } =-1$ is represented by (1), then
$m{ \left( -1 \right)  }^{ 2 }+\left( 1-{ m }^{ 2 } \right) \left( -1 \right) -m=0\Rightarrow 1-{ m }^{ 2 }=0\Rightarrow m=\pm 1$
In both cases either $m=1$ or $m=-1.$

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

If the pair of straight lines ${x^2} - 2pxy - {y^2} = 0$ and ${x^2} - 2qxy - {y^2} = 0$ be such that each pair bisects the angle between the other pair,then:

  1. $pq=-1$
  2. $p=q$
  3. $p=-q$
  4. $pq=1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If each pair bisects the angle between the other, the lines must be the same, which implies the coefficients must match, leading to p = q.

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

The pairs of straight lines $ax^{2}+2hxy-ay^{2}=0$ and $hx^{2}-2axy-hy^{2}=0$ are such that

  1. one pair bisects the angles between the other pair

  2. the lines of one pair are equally inclined to the lines of the other pair

  3. the lines of one pair are perpendicular to the `lines of the other pair

  4. none of these
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Given pairs of straight lines are $ax^{2}+2hxy-ay^{2}=0$ and $hx^{2}-2axy-hy^{2}=0$.

Pair of angular bisectors of $ax^{2}+2hxy-ay^{2}=0$ is $h(x^2-y^2)=(a-(-a))xy$

$\Rightarrow hx^2-2axy-hy^2$

$\therefore$ One pair bisects the angle between the other.

Clearly, if one pair bisects the angle between the other, the lines of one pair are equally inclined to the lines of the other pair.
Hence, option A and B.

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

If one of the lines of $my^2 + (1- m^2) xy - mx^2 = 0$ is a bisector of the angle between the lines $xy = 0$, then $m$ is

  1. $1$
  2. $2$
  3. $\displaystyle \frac{-1}{2}$
  4. $-1$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

Angle bisectors of $xy=0$ are $x+y=0 $  and $x=y$
$my^2+xy-m^2 xy -mx^2=0$
$\therefore y(my+x)-mx(my+x)=0$
$\therefore (y-mx)(my+x)=0$
Comparing $y-mx=0  $ and $  my+x=0  $  with  $y=x $ and  $ y=-x$,  we get $m=\pm 1$ 

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

The straight lines $7x^{2}+6xy+4y^{2}=0$ have the same pair of bisectors as those of the lines given by

  1. $49x^{2}+66xy+16y^{2}=0$
  2. $10x^{2}+6xy+7y^{2}=0$
  3. $5x^{2}+6xy+2y^{2}=0$
  4. $4x^{2}-6xy+7y^{2}=0$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

For $7x^{ 2 }+6xy+4y^{ 2 }=0$ 
Equation of angle bisector is 
$\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ 7-4 } =\cfrac { xy }{ 3 } \Rightarrow { x }^{ 2 }-xy-{ y }^{ 2 }=0$
For Option A 
Equation of angle bisector of $49x^{ 2 }+66xy+16y^{ 2 }=0$ is
$\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ 33 } =\cfrac { xy }{ 33 } \Rightarrow { x }^{ 2 }-xy-{ y }^{ 2 }=0$
For Option B
Equation of angle bisector of $10x^{ 2 }+6xy+7y^{ 2 }=0$ is
$\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ 3 } =\cfrac { xy }{ 3 } \Rightarrow { x }^{ 2 }-xy-{ y }^{ 2 }=0$
For Option C
Equation of angle bisector of $5x^{ 2 }+6xy+2y^{ 2 }=0$ is
$\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ 3 } =\cfrac { xy }{ 3 } \Rightarrow { x }^{ 2 }-xy-{ y }^{ 2 }=0$
For Option D
Equation of angle bisector of $4x^{ 2 }-6xy+7y^{ 2 }=0$ is
$\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ -3 } =\cfrac { xy }{ -3 } \Rightarrow { x }^{ 2 }-xy-{ y }^{ 2 }=0$

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines
$\displaystyle ax^{2}+2hxy+by^{2}=0$ represents a pair of straight lines through origin & angle between them is given by
$\displaystyle \tan \theta=\frac{2\sqrt{h^{2}-ab}}{a+b}$. If the lines are perpendicular then $\displaystyle a+b=0 $ and the equation of bisectors is given by  $\displaystyle \frac{x^{2}-y^{2}}{a-b}=\frac{xy}{h}$
The general equation of second degree given by
$\displaystyle ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ represent a pair of straight lines if $\displaystyle \triangle =0 $ or 
$ \displaystyle \begin{vmatrix}a&h  &g \\ h&b  &f \\ g&f  &c \end{vmatrix}=0 $ or $\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$
On the basis of above information answer the following question

Let $\displaystyle  f _{1}\left (x,y  \right )=ax^{2}+2hxy+by^{2}=0$ and let $\displaystyle  f _{i+1}\left (x,y  \right )=0 $ denotes the equation of bisectors of $\displaystyle  f _{i}\left (x,y  \right )=0 \forall $ $ i=1,2,3 $ then equation of $\displaystyle  f _{3}\left (x,y  \right )=0$ is

  1. $\displaystyle \left (a-b \right )x^{2}-4hxy+\left (a-b \right )y^{2}=0 $
  2. $\displaystyle \left (a-b \right )x^{2}-4hxy-\left (a-b \right )y^{2}=0 $
  3. $\displaystyle \left (a-b \right )x^{2}+4hxy-\left (a-b \right )y^{2}=0 $
  4. $\displaystyle \left (a-b \right )x^{2}+4hxy+\left (a-b \right )y^{2}=0 $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The bisectors of the lines f1(x,y) = 0 are given by f2(x,y) = 0. The bisectors of the bisectors f2(x,y) = 0 are the original lines f1(x,y) = 0 (or a rotated version). Iterating this process leads to the result.

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

The sum and product of the slopes of a pair of straight lines are the arithmetic and the geometric means of 9 and 16 respectively. The equation of the bisectors of the angles between the lines through the origin are 

  1. $24x^{2}-25xy+2y^{2}=0$
  2. $25x^{2}+44xy-25y^{2}=0$
  3. $11x^{2}-25xy-11y^{2}=0$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the equation of lines passing through origin be $y-m _1x=0,y-m _2x=0$
$\therefore$The combined equation of pair of straight lines $=(y-m _1x)(y-m _2x)=0$

$\Rightarrow m _1m _2x^2-(m _1+m _2)xy+y^2=0$

But given sum of the slopes ,$m _1+m _2=\frac{(9+16)}{2}=\frac{25}{2}$
product of the slopes ,$m _1m _2=\sqrt(9*16)=12$
On subtituting these values in the above equation.
$\Rightarrow 12x^2-\displaystyle\frac{25}{2}xy+y^2=0$

$\Rightarrow 24x^2-25xy+2y^2=0$ comparing with general equation of pair of straight lines passing through origin $ax^2+2hxy+by^2=0$
$\Rightarrow a=224,h=\displaystyle\frac{-25}{2},b=2$
If $ax^2+2hxy+cy^2=0$ is pair of equation of line passing through origin then pair of equation of the angular bisector of these pair of lines is obtained by
 $h(x^2-y^2)=(a-b)xy$

$\therefore$ The required pair of equation of angular bisector is $h(x^2-y^2)=(a-b)xy$
$\Rightarrow \displaystyle\frac{-25}{2}(x^2-y^2)=(24-2)xy$
$\Rightarrow \displaystyle\frac{25}{2}*(x^2-y^2)=-22xy$

$\Rightarrow (25x^2-25y^2)=-44xy$

$\Rightarrow 25x^2+44xy-25y^2=0$

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

If $\displaystyle y=mx$ bisects the angle between the lines $\displaystyle x^{2}\left ( \tan ^{2}\theta +\cos ^{2}\theta  \right )+2xy\tan \theta -y^{2}\sin ^{2}\theta =0$  when $\displaystyle \theta =\dfrac\pi3$ the value of $m$ is

  1. $\displaystyle \frac{-2- \sqrt 7}{ \sqrt 3}$
  2. $\displaystyle \frac{ \sqrt 7-2}{ \sqrt 3}$
  3. $\displaystyle 2 \sqrt 7 $
  4. $\displaystyle 2 \sqrt 3 $
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation
Equation of the bisectors of the angles between the given lines is

$\displaystyle \dfrac{x _2-y _2}{a-b}=\dfrac{xy}{h }$

Equation of the bisectors of the angles between the given lines is
$\displaystyle \frac{x^{2}-y^{2}}{\tan ^{2}\theta +\cos ^{2}\theta +\sin ^{2}\theta }=\frac{xy}{\tan \theta }$

$\displaystyle \Rightarrow \frac{x^{2}-y^{2}}{1+\tan ^{2}\theta  }=\frac{xy}{\tan \theta }$

$\displaystyle \Rightarrow \frac{x^{2}-y^{2}}{1+3  }=\frac{xy}{\sqrt 3 } \ when \ \ \theta=\pi/3$

Which satisfied by $y=mx $ if

$\displaystyle \frac{1-m^{2}}{4}=\frac{m}{\sqrt 3}$

$\displaystyle \Rightarrow \sqrt 3 m^{2}+4m-\sqrt 3=0$

$\displaystyle \Rightarrow m=\frac{-2\pm \sqrt 7}{\sqrt 3}$
Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

If two of the lines represented by $ x^{4} + x^{3} y + cx^{2}y^{2} -xy^{3} + y^{4} =0$ bisect the angle between the other two, then the value of $c$ is

  1. $0$
  2. $-1$
  3. $1$
  4. $-6$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since the product of the slopes of the four lines represented by the given equation is $1$ and a pair of lines represent the bisectors of the angles between the other two, the product of the slopes of each pair is $-1$. So let the equation of one pair be $ax^{2} + 2hxy -ay^{2} = 0$

The equation of its bisectors is $ \displaystyle \frac{x^{2}-y^{2}}{2a}=\frac{xy}{h} $

By hypothesis $ x^{4} +x^{3}y + cx^{2} y^{2}-xy^{3} + y^{4} $ $= (ax^{2} + 2hxy -ay^{2}) (hx^{2} -2axy -hy^{2})$ 

$ = ah(x^{4} + y^{4}) + 2(h^{2} -a^{ 2}) (x^{3}y- xy^{3}) -6ahx^{2}y^{2} $ 

Comparing the respective coefficients we get

$ah = 1 $ and $c = -6ah = -6$

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

The line $y=3x$ bisects the angle between the lines $ax^{2}+2axy+y^{2}=0$ if ${a}=$ 

  1. $3$
  2. $11$
  3. $\displaystyle \frac{3}{11}$
  4. $\displaystyle \frac{11}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Pair of angle bisectors represented by
$ax^2+2hxy+by^2=0$   is given by
$\dfrac{x^2-y^2}{a-b}=\dfrac{xy}{h}$
$\therefore ax^2+2axy+y^2=0$
pair of angle bisector is,
$\dfrac{x^2-y^2}{a-1}=\dfrac{xy}{a}$
$ax^2-ay^2=(a-1)xy$
Given  $ y=3x$  is one of angle bisector of given lines
$m=\dfrac{y}{x},$        $am^2+(a-1)x-a=0$
$m=3$ is satisfied to this equation
$9a+3a-3-a=0$
$11a=3$
$\therefore a=\dfrac{3}{11}$

Multiple choice line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

The direction cosines of a line parallel to the planes $\displaystyle 3x + 4y + z = 0$ and $\displaystyle x - 2y - 3z = 5$ are

  1. $\displaystyle \left ( -1, \: 1, \: -1 \right )$
  2. $\displaystyle \left ( -\frac{1}{\sqrt{3}}, \: -\frac{1}{\sqrt{3}}, \: \frac{1}{\sqrt{3}} \right )$
  3. $\displaystyle \left ( -\frac{1}{\sqrt{3}}, \: \frac{1}{\sqrt{3}}, \: \frac{-1}{\sqrt{3}} \right )$

  4. no line possible
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given equations of the planes are $\displaystyle 3x + 4y + z = 0$ and $\displaystyle x - 2y - 3z = 5$
required line is parallel to the given palnes,i.e perpendicular to the normals to the planes whose direction ratios are
$(3,4,1)$ and $(1,-2,-3)$ respectively
let $(a,b,c)$ be direction ratios of the line.
$\Rightarrow 3a+4b+c=0 and a-2b-3c=0$
$\Rightarrow a=-b=c$
$\therefore$ direction cosines of the line are $\displaystyle \left ( -\frac{1}{\sqrt{3}}, : \frac{1}{\sqrt{3}}, : -\frac{1}{\sqrt{3}} \right )$ or $\displaystyle \left ( \frac{1}{\sqrt{3}}, : -\frac{1}{\sqrt{3}}, : \frac{1}{\sqrt{3}} \right )$   

Multiple choice line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

The direction ratios of the line $x-y+z-5=0=x-3y-6$ are 

  1. $3,1,-2$
  2. $2,-4,1$
  3. <p class="MsoNormal">$\displaystyle \dfrac { 3 }{ \sqrt { 14 } } ,\dfrac { 1 }{ \sqrt { 14 } } ,\dfrac { -2 }{ \sqrt { 14 } } $</p>
  4. <p class="MsoNormal">$\displaystyle \dfrac { 2 }{ \sqrt { 14 } } ,\dfrac { -4 }{ \sqrt { 14 } } ,\dfrac { 1 }{ \sqrt { 14 } } $</p>
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If $l,m,n$ are the d.c's of the line, then

$1.l-1.m+1.n=0$

and $1/l-3.m+0.n=0$

$\displaystyle \therefore \dfrac { l }{ 0+3 } +\dfrac { m }{ 1-0 } =\dfrac { n }{ -3+1 } $

Hence, the dr's of the line are $3,1,-2$.

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

If the equation $a{x}^{2}+2hxy+b{y}^{2}=0$ represents a pair of lines then  the equation of the pair of lines of angular bisectors is $h({x}^{2}-{y}^{2})-(a-b)xy=0$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation of the pair of angular bisectors for the lines represented by ax^2 + 2hxy + by^2 = 0 is indeed given by (x^2 - y^2)/h = xy/(a - b), which rearranges to h(x^2 - y^2) - (a - b)xy = 0.

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

If the line $y = mx$ bisects the angle between the line $ax^2 + 2h\ xy + by^2 = 0$ then $m$ is a root of the quadratic equation :

  1. $hx^2 + (a - b)x - h = 0$
  2. $x^2 +h(a - b)x - 1 = 0$
  3. $(a - b)x^2 + hx - (a - b) = 0$
  4. $(a - b)x^2 - hx - (a - b) = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation of bisectors of the pair of straight lines $ax^2+2hxy+by^2=0$ is

$h(x^2-y^2)-(a-b)xy=0$......(1).

Since $y=mx $ is given to be the bisector of the pair of straight lines, then the line will satisfy the equation (1).

Then we get,
$h(1-m^2)-(a-b)m=0$

$hm^2+(a-b)m-h=0$.

So $m$ satisfies the equation $hx^2+(a-b)x-h=0$.

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

The equation of the bisector of the obtuse angle between the lines 3x-4y+7=0 and 12x+5y-2=0 is: 

  1. 21 x+77y-101=0

  2. 21 x+77 y+101=0

  3. 21x-77y-101=0

  4. 21x-77y+101=0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To find the bisector of the obtuse angle, calculate the bisectors using the formula (3x - 4y + 7)/5 = +/- (12x + 5y - 2)/13. Test which one corresponds to the obtuse angle by checking the sign of the expression.