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 vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

The length of the perpendicular drawn from $(1, 2, 3)$ to the line $\dfrac {x-6}{3}=\dfrac {y-7}{2}=\dfrac {z-7}{-2}$ is-

  1. $4$
  2. $5$
  3. $6$
  4. $7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let us take a point on line $(3\lambda +6,2\lambda +7,-2\lambda +7)$.
Direction ratio's of line which is  perpendicular to given line 
$(3\lambda +6-1,2\lambda +7-2,-2\lambda +7-3)$
$(3\lambda +5,2\lambda +5,-2\lambda +4)$
and the direction ratio's of given line are 3,2,-2
These two lines are perpendicular, so$(3\lambda +5)*3+(2\lambda +5)*2+(-2\lambda +4)+(-2)=0$
$\lambda=-1$
So, point is $(2,4,6)$
So distance between $(3,5,9)$ and $(1,2,3)$ is $7$. (by distance formula)

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

The length of the perpendicular from (1,6,3) to the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ is 

  1. 3
  2. $\sqrt{11}$
  3. $\sqrt{13}$

  4. 5
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Line: x=k, y=2k+1, z=3k+2. Vector from (1,6,3) to (k, 2k+1, 3k+2) is (k-1, 2k-5, 3k-1). Dot product with (1, 2, 3) = 0 => k-1 + 4k-10 + 9k-3 = 0 => 14k = 14 => k=1. Point is (1, 3, 5). Distance = sqrt((1-1)^2 + (6-3)^2 + (3-5)^2) = sqrt(0 + 9 + 4) = sqrt(13).

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

If $(a-a\prime )^2+(b-b\prime )^2+(c-c\prime )^2=p$ and $(ab\prime -a\prime b)^2+(bc\prime -b\prime c)^2+(ca\prime -c\prime a)^2=q,$ then the perpendicular distance of the line $ax+by+cz=1,$ $a\prime x+b\prime y+c\prime z=1$ from origin, is 

  1. $\sqrt { \dfrac { p }{ q } } $
  2. $\sqrt { \dfrac { q }{ p } } $
  3. $\dfrac { p }{ \sqrt { q } } $
  4. $\dfrac { q }{ \sqrt { p } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The distance of the intersection line of two planes from the origin involves the coefficients of the planes. The given expressions p and q relate to the distance formula for the line of intersection.

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

Perpendiculars AP, AQ and AR are drawn to the $x-,y-$ and $z-$axes, respectively, from the point $A\left ( 1,-1,2 \right )$. The A.M. of $AP^2,$ $AQ^2$ and $AR^2$ is

  1. $4$
  2. $5$
  3. $3$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$A=(1,-1,2), P=(1,0,0), ,Q=(0,-1,0),R=(0,0,2)$
$AP^2 = [(-1)^2+2^2] = 5$,
$ AQ^2  = [1^2+2^2] = 5$,
$ AR^2 = [(-1)^2+1^2] = 2$
Hence required A.M is $=\cfrac{5+5+2}{3}=4$ 
Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

The length of the perpendicular drawn from $(1,2,3)$ to the line $\displaystyle \frac { x-6 }{ 3 } =\frac { y-7 }{ 2 } =\frac { z-7 }{ -2 } $ is

  1. $4$
  2. $5$
  3. $6$
  4. $7$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Direction cosines of the given line are  

$\displaystyle \frac { 3 }{ \sqrt { 17 }  } ,\frac { 2 }{ \sqrt { 17 }  } ,\frac { -2 }{ \sqrt { 17 }  } $
$\displaystyle \therefore AM=\left| \left( 6.1 \right) .\frac { 3 }{ \sqrt { 17 }  } +\left( 7-2 \right) .\frac { 2 }{ \sqrt { 17 }  } +\left( 7-3 \right) .\frac { -2 }{ \sqrt { 17 }  }  \right| =17$
$AP=\sqrt { { \left( 16-1 \right)  }^{ 2 }+{ \left( 7-2 \right)  }^{ 2 }+{ \left( 7-3 \right)  }^{ 2 } } $
$=\sqrt { 25+25+16 } =\sqrt { 66 } $
$\therefore$ length of the perpendicular is
$PM=\sqrt { { AP }^{ 2 }-{ AM }^{ 2 } } $
$=\sqrt { 66-17 } =\sqrt { 49 } =7$

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

The distance of the point $A(-2,3,1)$ from the line $BC$ passing through $B(-3,5,2)$ which makes equal angles with the axes is

  1. $\displaystyle \dfrac{2}{\sqrt{3}}$
  2. $\sqrt{\dfrac{14}{3}}$
  3. $\displaystyle \dfrac{16}{\sqrt{3}}$
  4. $\displaystyle \dfrac{5}{\sqrt{3}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since $ \alpha = \beta = \gamma $
$\Rightarrow l = m = n = \dfrac{1}{\sqrt{3}}$, where $l,m,n$ are direction cosines of line $PQ$
Let $M$ be a point on the line $PQ$ such that $AM \perp PQ$
So, $PM$  $ =$   Projection of $AP$ on $ PQ$
          

 $ = \mid (-2 + 3)\dfrac{1}{\sqrt{3}} + (3 - 5)\dfrac{1}{\sqrt{3}} +(1 - 2)\dfrac{1}{\sqrt{3}} \mid = \dfrac{2}{\sqrt{3}}$

and $ AP= \sqrt{(-2+3)^2 + (3-5)^2 + (1-2)^2} = \sqrt{6}$
Hence required distance is,
$AM = \sqrt{PQ^2 - QM^2} = \sqrt{\dfrac{14}{3}} $

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

Find the length of perpendicular from $ P(2, -3, 1)$ to the line $\displaystyle \frac{x- 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{-1}$

  1. $5$
  2. $\displaystyle \sqrt{\dfrac{531}{14}}$
  3. $\sqrt{50}$
  4. $\sqrt{\dfrac{221}{3}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let take a point on line $(2\lambda +1,3\lambda +3,-\lambda -2)$


Direction ratio's of line which is  perpendicular to given line, 

$(2\lambda +1-2,3\lambda +3+3,-\lambda -2-1)$

$(2\lambda -1,3\lambda +6,-\lambda -3)$

And the direction ratio's of given line are $2,3,-1$.

These two lines are perpendicular, so

$(2\lambda -1)\cdot 2+(3\lambda +6)\cdot 3+(-\lambda -3)+(-1)=0$


$\lambda=\dfrac{-19}{14}$

So point is $\left (\dfrac{-24}{14},\dfrac{-15}{14},\dfrac{-9}{14}\right)$

So distance between $\left (\dfrac{-24}{14},\dfrac{-15}{14},\dfrac{-9}{14}\right) $ and $(1,3,-2)$ is $\sqrt{\dfrac{531}{14}}$...................(by distance formula) 

Multiple choice maths vectors: lines in two and three dimensions distance from a point to line perpendicular distance of a point from a plane the distance from a point to a line

A line is drawn from $P(x _1 , y _1)$ in the direction $\theta$ with the X - axis, to meet $ax + by + c = 0$ at $Q$. Then length $PQ$ is equal to :

  1. $\dfrac{|ax _1 + by _1 + c|}{\sqrt{(a^2 + b^2)}}$
  2. $\left|\dfrac{ax _1 + by _1 + c}{a \, cos \theta + b \, sin \theta} \right|$
  3. $\dfrac{bx _1 + ay _1 + c}{a cos \theta + b sin \theta}$
  4. $ - \dfrac{ax _1 + by _1 + c}{a sin \theta + b cos \theta}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Equation of a line drawn through a point $P\left( { x } _{ 1 }{ y } _{ 1 } \right) $ at an angle $\theta $ with the $x-axis$
$\dfrac { x-{ x } _{ 1 } }{ \cos\theta  } =\dfrac { y-{ y } _{ 1 } }{ \sin\theta  } \quad \longrightarrow \left( 1 \right) $
$Q$ is a point the above line and also lies on the line $ax+by+cz=0$
Say $\left| PQ \right| =r$ and the coordinates of $Q$ are $\left( h,k \right) $
Then,
$h={ x } _{ 1 }+r\cos\theta $
$k={ y } _{ 1 }+r\sin\theta $
$\because$   $Q$ lies on $ax+by+cz=0$
$\therefore$   $ah+bk+c=0$
$\Rightarrow a\left( { x } _{ 1 }+r\cos\theta  \right) +b\left( { y } _{ 1 }+r\sin\theta  \right) +c=0$
$\Rightarrow { ax } _{ 1 }+{ by } _{ 1 }+c+r\left( a\cos\theta +b\sin\theta  \right) =0$
$\Rightarrow \quad r=\dfrac { -\left( { ax } _{ 1 }+{ by } _{ 1 }+c \right)  }{ a\cos\theta +b\sin\theta  } $
$\because$   $'r'$ is the magnitude of length of line segment $PQ$, it cannot be negative.
So, $r=\left| \dfrac { { ax } _{ 1 }+{ by } _{ 1 }+c }{ a\cos\theta +b\sin\theta  }  \right| $
Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

The condition that the line $\displaystyle \frac{x-{\alpha }'}{l}=\frac{y -{\beta   }'}{m}=\frac{z-{\gamma  }'}{n}$ in the plane $Ax + By + Cz + D = 0$ is

  1. $A{\alpha }'+B{\beta }'+C{\gamma }'+D=0\ and\ Al+Bm+Cn\neq 0$
  2. $A{\alpha }'+B{\beta }'+C{\gamma }'+D\neq0\ and\ Al+Bm+Cn= 0$
  3. $A{\alpha }'+B{\beta }'+C{\gamma }'+D=0\ and\ Al+Bm+Cn= 0$
  4. $A{\alpha }'+B{\beta }'+C{\gamma }'=0\ and\ Al+Bm+Cn= 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The line $\dfrac{x- \alpha^1}{l}=\dfrac{y-\beta^1}{m}=\dfrac{z-\gamma^1}{n}$ in the plane $Ax+By+Cz+D=0$

then multiplication sum id corresponding direction ratio will be zero so here $Al+Bm+Cn=0$
 and  one thing more line passes through $(\alpha^1,\beta^1,\gamma^1)$
So, plane Also passe through $(\alpha^1,\beta^1,\gamma^1)$
hence $A\alpha^1+B\beta^1+C\gamma^1+D=0$

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

If a line which passes through the point $A(0,\,1,\,2)$ and makes angle $\displaystyle\frac{\pi}{4},\,\displaystyle\frac{\pi}{4},\,\displaystyle\frac{\pi}{2}$ with $x,\,y,\,&\,z$ axes respectively. The line meets the plane $x+y+z=0$ at point $B$. The length $\sqrt{2}AB$ is equal to

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

The D.C. of the line are $\begin{pmatrix}\displaystyle\frac{1}{\sqrt{2}},\,\displaystyle\frac{1}{\sqrt{2}},\,0\end{pmatrix}$
any point on the line at a distance $\lambda$ from $A(0,\,1,\,2)$
is $\begin{pmatrix}0+\displaystyle\frac{\lambda}{\sqrt{2}},\,1+\displaystyle\frac{\lambda}{\sqrt{2}},\,2+\lambda.0\end{pmatrix}$
which lies on $x+y+z=0$
$\therefore\;\displaystyle\frac{\lambda}{\sqrt{2}}+\begin{pmatrix}1+\displaystyle\frac{\lambda}{\sqrt{2}}\end{pmatrix}+2=0$
$\Rightarrow\;\displaystyle\frac{2\lambda}{\sqrt{2}}=-3\;\;\;\;\Rightarrow\;\;\;\;\lambda=\displaystyle\frac{-3}{\sqrt{2}}$
$\therefore\;B=\begin{pmatrix}-\displaystyle\frac{3}{2},\,-\displaystyle\frac{1}{2},\,2\end{pmatrix}$
$A=(0,\,1,\,2)\;&\;B\equiv\;\begin{pmatrix}-\displaystyle\frac{3}{2},\,-\displaystyle\frac{1}{2},\,2\end{pmatrix}$
$AB=\sqrt{\displaystyle\frac{9}{4}+\displaystyle\frac{9}{4}}=\displaystyle\frac{3\sqrt{2}}{2}=\displaystyle\frac{3}{\sqrt{2}}$

Multiple choice maths three dimensional geometry - ii point of intersection of a line and a plane line and a plane three dimensional geometry

A line with positive direction cosines passes through the point $\displaystyle P\left ( 2,-1,2 \right )$ and makes equal angles with the coordinates axis. The line meet the plane $\displaystyle 2x+y+z=9$ at ponit $Q$.
The length of the line segment $PQ$ equals.

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

Equation of a line through the point $\displaystyle P\left ( 2,-1,2 \right )$, equally inclined to the axes is given by,
$\displaystyle \frac{x-2}{1}=\frac{y+1}{1}=\frac{z-2}{1}=k$ (say)
Any point on the line is $Q ( k+2,k-1, k+2  )$ which lies on the plane $\displaystyle 2x+y+z=9$
 $\Rightarrow \displaystyle 2\left ( k+2 \right )+k-1+k+2=9 \Rightarrow k=1$,
For this values of $k$, the coordinates of $Q$ are $\displaystyle \left ( 3,0,3 \right )$.
So, $\displaystyle PQ=\sqrt{\left ( 3-2 \right)^2+\left ( 0+1 \right )^2+\left ( 3-2 \right )^2}=\sqrt{3}$

Multiple choice maths mixture types of ratios ratios in proportion mathematical logic

If a line with direction ratio $2:2:1$ intersects the line $\dfrac {x-7}{3}=\dfrac {y-5}{2}=\dfrac {z-3}{1}$ and $\dfrac {x-1}{2}=\dfrac {y+1}{4}=\dfrac {z+1}{3}$ at $A$ and $B$ then $AB=$

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

$\begin{array}{l} \frac { { x-7 } }{ 3 } =\frac { { y-5 } }{ 2 } =\frac { { z-3 } }{ 1 } \, \, \, \, \, \, \, \, Let\, \, A=\left( { 3{ r _{ 1 } }+7,\, \, 2{ r _{ 1 } }+5,\, \, { r _{ 1 } }+3 } \right)  \ \frac { { x-1 } }{ 2 } =\frac { { y+1 } }{ 4 } =\frac { { z+1 } }{ 3 } \, \, \, \, \, and,\, Let\, B=\left( { 2{ r _{ 2 } }+1,\, \, 4{ r _{ 2 } }-1,\, \, 3{ r _{ 2 } }-1 } \right)  \ Direction\, ratios\, of\, AB=\left( { 2{ r _{ 2 } }-3{ r _{ 1 } }-6,\, \, \, 4{ r _{ 2 } }-2{ r _{ 1 } }-6,\, \, \, 3{ r _{ 2 } }-{ r _{ 1 } }-4 } \right)  \ \frac { { 2{ r _{ 2 } }-3{ r _{ 1 } }-6 } }{ 2 } =\frac { { 4{ r _{ 2 } }-2{ r _{ 1 } }-6 } }{ 2 } =\frac { { 3{ r _{ 2 } }-{ r _{ 1 } }-4 } }{ 1 }  \ { r _{ 1 } }+2{ r _{ 2 } }=0\, \, \, \, \therefore { r _{ 1 } }=-2 \ and, \ 4{ r _{ 1 } }-2{ r _{ 1 } }-6=6{ r _{ 2 } }-2{ r _{ 1 } }-8 \ 2=2{ r _{ 2 } } \ \therefore { r _{ 2 } }=1 \ A\equiv \left( { 1,\, 1,\, 1 } \right) \, \, \, \, \, B=\left( { 3,\, 3,\, 2 } \right)  \ AB=\sqrt { 4+4+1 }  \ =3 \ Hence,\, Option\, D\, is\, the\, correct\, answer. \end{array}$

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

The axis of the conic $\displaystyle x^{2}+4y-6x+17=0$ is

  1. $\displaystyle x=5 $
  2. $\displaystyle y=5 $
  3. $\displaystyle x=3 $
  4. $\displaystyle x=-3 $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x^2 + 4y -6x +17 =0$

$\Rightarrow x^2 -6x = -4y-17$
$\Rightarrow x^2 - 6x + 9 = -4y + 8$
$\Rightarrow (x-3)^2 = -4(y-2)$
Let $X = x-3$ and $Y = y-2$, then we get
$X^2 = -4Y$
Comparing the above equation with the standard equation of the parabola, we get that the axis of the parabola is given by $X=0$
Hence, axis of the given parabola will be $x-3 = 0\Rightarrow x =3$
Option $C$ is correct.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Find the point of intersection and the inclination of the two lines $Ax+By=A+B$ and $A(x-y)+B(x + y)=2B$.

  1. $(1,1); 45^0$
  2. $(1,2), 60^0$
  3. $(2,1), 75^0$

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

$Ax+By=A+B$    ........(i)

$  A(x-y)+B(x+y)=2B$

$\Rightarrow  (A+B)x+(B-A)y=2B$    .......(ii)

$\Rightarrow  (A+B)x=2B-(B-A)y\\ \Rightarrow x=\dfrac { 2B-(B-A)y }{ A+B } $

Substituting $x$ in (i), we get

$A\left( \dfrac { 2B-(B-A)y }{ A+B }  \right) +By=A+B$

$\Rightarrow \dfrac { 2AB-ABy+{ A }^{ 2 }y+{ B }^{ 2 }y+ABy }{ A+B } =A+B$

$\Rightarrow 2AB-ABy+{ B }^{ 2 }y+{ A }^{ 2 }y+ABy={ A }^{ 2 }+{ B }^{ 2 }+2AB$ 

$\Rightarrow (A^{2}+{ B }^{ 2 })y=A^{2}+{ B }^{ 2 }$

$\Rightarrow y=1 $

Substituting $y$ in $(i)$

$\Rightarrow  Ax+B\left( 1  \right) =A+B\\ \Rightarrow Ax+B=A+B\\ \Rightarrow Ax=A\\ \Rightarrow x=1 $

So, the point of intersection is $(1,1) $.

Slope of (i), ${ m } _{ 1 }=-\dfrac { A }{ B } $.

Slope of (ii), ${ m } _{ 2 }=-\dfrac { (A+B) }{ B-A } =\dfrac { A+B }{ A-B } $

$\tan { \theta  } =\dfrac { { m } _{ 1 }-{ m } _{ 2 } }{ 1+{ m } _{ 1 }{ m } _{ 2 } } \\ \Rightarrow \tan { \theta  } =\dfrac { -\dfrac{A}{B}-\dfrac { A+B }{ A-B }  }{ 1-\dfrac{A}{B}\times\dfrac { A+B }{ A-B }  } $

$ \tan { \theta  } =-\dfrac { \left\{ \dfrac { { A }^{ 2 }+AB-AB+{ B }^{ 2 } }{ B(A-B) }  \right\}  }{ \left\{ \dfrac { { -B }^{ 2 }+AB-{ A }^{ 2 }-AB }{ B(A-B) }  \right\}  } \\ \tan { \theta  } =-\dfrac { { A }^{ 2 }+{ B }^{ 2 } }{ { -(A }^{ 2 }+{ B }^{ 2 }) } =1$

$\Rightarrow \theta=45^{o}$