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

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

  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. $\displaystyle x=\pm 2y$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice maths construction of polygons construction of parallelograms and rectangles construction of special quadrilaterals constructions related to a quadrilateral

If one diagonal of a square is the portion of the line $\frac { x }{ a } +\frac { y }{ b } =1$ intercepted by the axes, then the extremities of the other diagonal of the square are

  1. $\left( \frac { a+b }{ 2 } ,\frac { a+b }{ 2 } \right) $
  2. $\left( \frac { a-b }{ 2 } ,\frac { a+b }{ 2 } \right) $
  3. $\left( \frac { a-b }{ 2 } ,\frac { b-a }{ 2 } \right) $
  4. $\left( \frac { a+b }{ 2 } ,\frac { b-a }{ 2 } \right) $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The line x/a + y/b = 1 intercepts the axes at (a, 0) and (0, b). These are the extremities of one diagonal of a square. The center of the square is the midpoint ((a/2), (b/2)). The other diagonal is perpendicular and of equal length, leading to the coordinates ((a-b)/2, (b-a)/2) for the other extremities.

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 is one of the bisector of the lines $x^2 + 4xy - y^2 = 0$, then the value of no ___________.

  1. $\frac{\sqrt{5} - 1}{2}$
  2. $\frac{\sqrt{5} + 1}{2}$
  3. $-(\frac{\sqrt{5} + 1}{2})$
  4. $-(\frac{\sqrt{5} -1}{2})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The angle bisectors of the pair of lines ax^2 + 2hxy + by^2 = 0 are given by (x^2 - y^2)/(a - b) = xy/h. Substituting the coefficients from x^2 + 4xy - y^2 = 0, we find the slopes of the bisectors.

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

The Straight lines represented by the equation $135{ x }^{ 2 }-136xy+33{ y }^{ 2 }=0$ are equally inclined to the line 

  1. $x-2y=7$
  2. $x+2y=7$
  3. $x-2y=4$
  4. $3x+2y=4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Give pair of lines is $135{ x }^{ 2 }-136xy+33{ y }^{ 2 }=0$   ...(1)


The equation of bisector of angles between pair of lines (1) is

$\displaystyle \frac { { x }^{ 2 }-{ y }^{ 2 } }{ a-b } =\frac { xy }{ h } \Rightarrow \frac { { x }^{ 2 }-{ y }^{ 2 } }{ 135-33 } =\frac { xy }{ -68 } $

$\Rightarrow 2{ x }^{ 2 }+3xy-2{ y }^{ 2 }=0\Rightarrow \left( x+2y \right) \left( 2x-y \right) =0$

One of the bisectors is $x+2y=0$ which is parallel to the line $x+2y=7$.

Hence, the line $x+2y=7$ is equally inclined to the given lines.

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

Given equations are $\displaystyle  x^{2}-2qxy-y^{2}=0 ...(1) $ $\displaystyle  x^{2}-2pxy-y^{2}=0 ...(2) $ Joint equation of angle bisector of the line (i) and (ii) are same $\displaystyle \therefore qx^{2}+2xy-qy^{2}=0....(3) $. 


Now (2).and (3) are same, taking the ratio of their coefficients


$\displaystyle \therefore \frac{1}{q}=\frac{-p}{1}\Rightarrow pq=-1$
Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

2x + y - 4 = 0 is a besector of angles between the lines a(x - 1) + b(y - 2) = 0, c(x - 1) + d(y - 2) = 0 the other angular bisector is _______________.

  1. x - 2y + 1 = 0

  2. x - 2y - 3 = 0

  3. x - 2y + 3 = 0

  4. x + 2y - 5 = 0
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
We have $a\left(x-1\right)+b\left(y-2\right)=0$       .....$(1)$
and $c\left(x-1\right)+d\left(y-2\right)=0$       .....$(2)$

Clearly $\left(1,2\right)$ lie on both the lines and hence $\left(1,2\right)$ is their point of intersection.

Both the bisectors will pass through $\left(1,2\right)$
One of the bisector is $2x+y-4=0$

Other bisector will be perpendicular to this bisector.
Hence its equation will be $x-2y=\lambda$

It passes through $\left(1,2\right)$
$\Rightarrow\,1-4=\lambda$
$\Rightarrow\,\lambda=-3$

Hence the equation is $x-2y=-3$ or $x-2y+3=0$
Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

The equations of the bisectors of that angle between the lines $x+2y-11=0,:3x+6y-5=0$ which contains the point $\left(1,-3\right)$ is 

  1. $3x=19$
  2. $3y=7$
  3. $3x=19$ and $3y=7$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The lines are x + 2y - 11 = 0 and 3x + 6y - 5 = 0. These are parallel lines. The bisector of parallel lines is a line parallel to them, equidistant from both. Calculating the midpoint of the perpendicular distance gives the result.

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

The line $L$ has intercepts $a$ and $b$ on the co-ordinate axes keeping the origin fixed, the co-ordinate axes are related through a fixed angle. If the same line has intercepts c and d then

  1. $ \displaystyle \frac{1}{a^{2}}+\frac{1}{c^{2}}= \frac{1}{b^{2}+d^{2}} $
  2. $ \displaystyle \frac{1}{a^{2}}+\frac{1}{b^{2}}= \frac{1}{c^{2}}+\frac{1}{d^{2}} $
  3. $ \displaystyle a^{2}+c^{2}= b^{2}+d^{2} $
  4. $ \displaystyle a^{2}+b^{2}= c^{2}+d^{2} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Suppose we state he coordinate axis in the anti-clockwise direction through an angle $\alpha$.
The equation of the line $\alpha$ with respect to old axes is $\displaystyle\frac{x}{a}+\frac{y}{b}=1$
In this equation replacing $x$ by $x\cos\alpha-y\sin\alpha$
The equation of the line with respect to new axes is
$\displaystyle\frac{x\cos\alpha-y\sin\alpha}{a}+\frac{x\sin\alpha+y\cos\alpha}{b}=1$
$\displaystyle\Rightarrow x\left( \frac { \cos { \alpha  }  }{ a } +\frac { \sin { \alpha  }  }{ b }  \right) +y\left( \frac { \cos { \alpha  }  }{ b } -\frac { \sin { \alpha  }  }{ a }  \right) =1$   ...(1)
The intercept mode by (1) on the co-ordinate axes are given as $c$ and $d$.
Therefore, $\displaystyle\frac{1}{c}=\frac { \cos { \alpha  }  }{ a } +\frac { \sin { \alpha  }  }{ b } $ and $\displaystyle\frac{1}{d}=\frac { \cos { \alpha  }  }{ b } -\frac { \sin { \alpha  }  }{ a }$
Squaring and adding, we get $ \displaystyle \frac{1}{c^{2}}+\frac{1}{d^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}} $

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

$P: x^{2}-y^{2}+2y-1=0$
$L: x+y=3$

Equation of the angle bisectors of the pairs of lines P is

  1. $xy-y=0$
  2. $xy-x=0$
  3. $xy=0$
  4. $xy+y=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$P:{ x }^{ 2 }-{ \left( y-1 \right)  }^{ 2 }=0$

$\Rightarrow x+y-1=0$ and $x-y+1=0$

Equation of the angle bisector is

$\cfrac { A _1x+B _1y+C _1 }{ \sqrt {A _1^2+B _1^2 }  } =\pm \cfrac { A _2x+B _2y+C _2 }{ \sqrt {A _2^2+B _2^2}  } $
$\cfrac { x+y-1 }{ \sqrt { 2 }  } =\pm \cfrac { x-y+1 }{ \sqrt { 2 }  } $

$\Rightarrow x=0$ or $y-1=0$

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

If pairs of lines $3x^{2}-2pxy-3y^{2}=0$ and $5x^{2}-2qxy-5y^{2}=0$ are such that each pair bisects the angle between the other pair, then $pq$ is equal to

  1. $-1$
  2. $-3$
  3. $-5$
  4. $-15$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given pairs 
$3x^2-2pxy-3y^2=0$----(1)

$5x^2-2qxy-5y^2=0$----(2)

Equation (1) can be written as 
$3(x^2-y^2)=2pxy$----(3)

Eq of angle bisector of eq (2)
$\dfrac{x^2-y^2}{5+5}=\dfrac{xy}{-q}$

$\dfrac{x^2-y^2}{10}=\dfrac{xy}{-q}$----(4)

Now as the given in question the eq of angle bisector of one pair bisects the other pair

So dividing eq (4) by (3)
$\dfrac{\dfrac{x^2-y^2}{10}}{3(x^2-y^2)}=\dfrac{\dfrac{xy}{-q}}{2pxy}$

$\dfrac{1}{30}=\dfrac{1}{-2pq}$

$pq=-15$ 
Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

Slope of a bisector of the angle between the lines $4x^{2}-16xy-7y^{2}=0$ is

  1. $\displaystyle \frac{11+\sqrt{377}}{16}$
  2. $\displaystyle \frac{11-\sqrt{377}}{16}$
  3. $\displaystyle \frac{-3+2\sqrt{3}}{7}$
  4. $\displaystyle \frac{-3-2\sqrt{3}}{7}$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation
Given pair 
$4x^2-16xy-7y^2=0$
On comparing given eq with $ax^2+2hxy+by^2=0$
$a=4,b=-7,h=-8$
Eq of pair of Angle bisector 
$\dfrac{x^2-y^2}{a-b}=\dfrac{xy}{h}$
$\dfrac{x^2-y^2}{11}=\dfrac{xy}{-8}$
$-8x^2+8y^2=11xy$
$8y^2-11xy-8x^2=0$
$y=\dfrac{-(-11x)\pm\sqrt{121x^2+256x^2}}{16}$

$y=\dfrac{11x\pm\sqrt{377x^2}}{16}$

$y=\dfrac{11x\pm\sqrt{377}x}{16}$

$y=\left (\dfrac{11\pm\sqrt{377}}{16}  \right )x$
Comparing above eq with $y=mx+c$
$m=\left (\dfrac{11+\sqrt{377}}{16}  \right )$ and $\left (\dfrac{11-\sqrt{377}}{16}  \right )$
Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

$P: 2x^{2}-axy+6y^{2}=0$
$Q: 3x^{2}-8xy+4y^{2}=0$
If the bisectors of the angles between the lines represented by $P$ and $Q$ are same, the value of $a$ is

  1. $8$
  2. $\dfrac{16}{3}$
  3. $32$
  4. $-16$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equation of angle bisector of $2x^{ 2 }-axy+6y^{ 2 }=0$ is

$\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ 2-6 } =\cfrac { xy }{ -a/2 } \Rightarrow a{ x }^{ 2 }-a{ y }^{ 2 }-8xy=0$   ----------(1)

And equation of angle bisector of $3x^{ 2 }-8xy+4y^{ 2 }=0$ is

$\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ 3-4 } =\cfrac { xy }{ -4 } \Rightarrow 4{ x }^{ 2 }-4{ y }^{ 2 }-xy=0$  -----------(2)

From (1) and (2), we get $a=32$

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

If the lines represented by $x^2-2pxy-y^2=0$ are rotated about the origin through an angle $\theta,$ one in clockwise direction and other in anti-clockwise direction, then the equation of the bisector of the angle between the lines in the new positions is

  1. $px^2+2xy-py^2=0$
  2. $px^2+2xy+py^2=0$
  3. $x^2-2pxy-y^2=0$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given eq 
$x^2-2pxy-y^2=0$
comparing above eq with general form of eq $ax^2+2hxy+by^2=0$
$a=1,h=-p,b=-1$
Now the line is rotated one in clockwise and other is anticlockwise so the both eq are replaced by each other and form the same eq as it was so here we finding eq of angle bisector by formula
$\dfrac{x^2-y^2}{a-b}=\dfrac{xy}{h}$
$\dfrac{x^2-y^2}{1+1}=\dfrac{xy}{-p}$
$\dfrac{x^2-y^2}{2}=\dfrac{xy}{-p}$
$(-p)(x^2-y^2)=2xy$
$-px^2+py^2=2xy$
$px^2+2xy-py^2=0$
 
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

The Joint equation of the bisectors of the angle between the lines represented by $\displaystyle ax^{2}+2hxy+by^{2}=0 $ is

  1. a pair of perpendicular lines

  2. a pair of parallel lines
  3. a pair of intersecting lines but not $\displaystyle \perp$er

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation of bisectors of angle between the pair of straight line $\displaystyle ax^{2}+2hxy+by^{2}=0$ is given by $\displaystyle \frac{x^{2}-y^{2}}{a-b}=\frac{xy}{h}$


$\displaystyle\Rightarrow hx^{2}-\left ( a-b \right )xy-hy^{2}=0$

Here coefficient of $x^{2}+$ coefficient of $y^{2}=h+(-h)=0$

$\displaystyle \therefore $ above equation represent the pair of $\perp$er lines

Hence choice (a) is correct.

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

$my^2+(1-m^2)xy-mx^2$
$(my+x)(y-mx)=0$
Therefore the lines are $y=mx$ and $y=\dfrac{-x}{m}$ ...(i)
The angle bisectors of the $xy=0$ is $y=x$ and $y=-x$
Comparing the slopes with the equations of the (i) we get
$m=\pm1$