Mathematics

Straight Lines and Coordinates

164 Questions

Straight lines and coordinates form the basis of coordinate geometry. This topic focuses on finding slopes, equations of lines, and points of intersection. These mathematical concepts are essential for performing well in advanced quantitative aptitude tests.

Line equations and slopesPoint of intersectionConcurrent linesNormal and parallel linesAngle between lines

Straight Lines and Coordinates Questions

Multiple choice maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

Find the equation of a line passing through the point (2, -3 ) and parallel to the line 2x - 3y + 8 = 0

  1. 2x - 3y =13

  2. 2x -3y = 12

  3. x - 3y =4

  4. 3x - 2y = 7
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation of the line parallel to $ 2x-3y+8 = 0 $ will be of the form $ 2x-3y + k = 0 $

Now, since it passes through $ (2,-3) $, on substituting it , we get $ 2(2) -3(-3) + k = 0  $
$ => k = -13 $

So, required eqn of parallel line is $ 2x - 3y - 13 = 0 $

Multiple choice maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

Which of the following is true about the three lines
$L _{1}: x - 3y + 7 = 0 , L _{2} : 2x + y - 3 = 0$ and $L _{3} : 7x +\dfrac{7y}{2}-\dfrac{21}{2}=0$

  1. Lines form a triangle

  2. Lines are concurrent

  3. Lines can not bound any region

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

Given equations of lines as
$L _{1}: x - 3y + 7 = 0 $
Slope of $L _{1}=\displaystyle \frac{1}{3}$
$ L _{2} : 2x + y - 3 = 0$ 
Slope of $L _{2}=-2$
$L _{3} : 7x +\dfrac{7y}{2}-\dfrac{21}{2}=0$
Slope of $L _{3}=-2$
$\Rightarrow L _{2}$ and $L _{3}$ are parallel
Hence, lines cannot bound any region.

Multiple choice maths sequences, functions and graphs reading graphs describing different situations using equations to plot lines basics of a straight line

Find the equation of the straight line passing through the point $ (6,2)  $ and having slope $ -3 .  $

  1. $x-3y-10=0$
  2. $3x+y-20=0$
  3. $x+2y-40=0$
  4. $3x-y-10=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation of any line is $y=mx+c$


here $m$ is slope of line 


so we have $m=-3$

$y=-3x+c$

Also this line passes through $(6, 2)$

$2=-18+c\Rightarrow c=20$

so equation of line will be $y+3x=20$

or $3x+y-20=0$.


Multiple choice reciprocal equations theory of equations maths

The equation of the line, reciprocal of whose intercepts on the axes are $a$ and $b$ given by

  1. $\dfrac x 2$ + $\dfrac yb$ = $1$
  2. $ax + by = 1$
  3. $ax + by = ab$
  4. $ax = by = 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let A & B be the part of intersection of line with X & Y axis respectively.

$\Rightarrow A= \left(\cfrac {1}{a},0\right)$
$\Rightarrow B= \left(0,\cfrac {1}{b}\right)$
$\therefore$ Equation of line= $\left(y-\cfrac {1}{b}\right)=\left(\cfrac {\cfrac {1}{b}-0}{0-\cfrac {1}{a}}\right)$
$\Rightarrow \left(\cfrac {1}{a}\right)\left(y-\cfrac {1}{b}\right)=\cfrac {1}{b}x$
$\Rightarrow \cfrac {-y}{a}+\cfrac {1}{ab}= \cfrac {x}{b}$
$\Rightarrow \cfrac {x}{b}+\cfrac {y}{a}=\cfrac {1}{ab}$
$\Rightarrow ax+by=1$

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

The equation of a line parallel to $x+2y=1$ and passing through the point of intersection of the lines $x-y=4$ and $3x+y=7$ is ?

  1. $x+2y=5$
  2. $4x+8y-1=0$
  3. $4x+8y+1=0$

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

Let the equation of line parallel to $x+2y=1$ be $y=mx+c$

intersection point of line $x-y=4$ and $3x+y=7$ is given by solving two equation we get $x=\dfrac { 11 }{ 4 } $ and $y=\dfrac { -5 }{ 4 } $
from equation $x+2y=1\ \Rightarrow y=-\dfrac { 1 }{ 2 } x+\dfrac { 1 }{ 2 } $
we get ${ m } _{ 1 }=-\dfrac { 1 }{ 2 } $
since $y=mx+c$ is parallel to $x+2y=1$
$\therefore { m } _{ 2 }=-\dfrac { 1 }{ 2 } $
also, $y=mx+c\ \Rightarrow \dfrac { -5 }{ 4 } =\dfrac { -1 }{ 2 } .\dfrac { 11 }{ 4 } +c\ \Rightarrow c=\dfrac { 1 }{ 8 } $
therefore equation of line parallel to $x+2y=1$ is given as 
$y=mx+c\ \Rightarrow y=\dfrac { -1 }{ 2 } x+\dfrac { 1 }{ 8 } \ \Rightarrow 4x+8y-1=0$

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

The line $x+y=1$ meets the lines represented by the equation $y^{3}-xy^{2}-14x^{2}y+24x^{3}=0$ at the points $A, B, C$. If $O$ is the origin, then $OA^{2}+OB^{2}+OC^{2}$ is equal to

  1. $\dfrac{22}9$
  2. $\dfrac{85}{72}$
  3. $\dfrac{181}{72}$
  4. $\dfrac{221}{72}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

X-coordinate of the points are given by the roots of the equation


$24{ x }^{ 3 }+14{ x }^{ 2 }\left( x-1 \right) -x{ \left( x-1 \right)  }^{ 2 }-{ \left( x-1 \right)  }^{ 3 }=0\ \Rightarrow 36{ x }^{ 3 }-9{ x }^{ 2 }-4x+1=0\ \Rightarrow \left( 3x-1 \right) \left( 3x+1 \right) \left( 4x-1 \right) =0\ \Rightarrow x=\cfrac { 1 }{ 3 } ,-\cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 4 } $

$\Rightarrow A\left( \cfrac { 1 }{ 3 } ,\cfrac { 2 }{ 3 }  \right) ,B\left( -\cfrac { 1 }{ 3 } ,\cfrac { 4 }{ 3 }  \right) $ and $C\left( \cfrac { 1 }{ 4 } ,\cfrac { 3 }{ 4 }  \right) $

Hence,

${ OA }^{ 2 }+{ OB }^{ 2 }+{ OC }^{ 2 }=\cfrac { 1 }{ 9 } +\cfrac { 4 }{ 9 } +\cfrac { 1 }{ 9 } +\cfrac { 16 }{ 9 } +\cfrac { 1 }{ 16 } +\cfrac { 9 }{ 16 } =\cfrac { 221 }{ 72 } $

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

The straight line passes through the point of intersection of the straight lines $x+2y-10=0$ and $2x+y+5=0$, is 

  1. $5x-4y=0$
  2. $5x+4y=0$
  3. $4x-5y=0$
  4. $4x+5y=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The line passes through the point of intersection of the equations $x+2y-10=0$  and $2x+y+5=0$

Now,
$\ x+2y-10=0....(i)\ 2x+y+5=0....(ii)\times 2\ =>4x+2y+10=0....(iii)$
Subtracting (iii) and (i), we get,
$-3x-20=0\ =>x=\cfrac { -20 }{ 3 } \ \therefore y=\cfrac { 25 }{ 3 } $
Now the line must pass through $(\cfrac { -20 }{ 3 } ,\cfrac { 25 }{ 3 } )$ 
Therefore, $5x+4y=0$ passes through $(\cfrac { -20 }{ 3 } ,\cfrac { 25 }{ 3 } )$

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

The lines $x+y=\left|\ a\ \right|$ and $ax-y=1$ intersect each other in the first quadrant. Then the set of all possible values of $a$ is the interval :

  1. $\left( 0,\infty \right)$
  2. $\left[ 1,\infty \right)$
  3. $\left( -1,\infty \right)$
  4. $\left( -1,1 \right] $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given lines are :

$x+y=\left | a\right |$ and $ax-y=1$
Case $1:a>0$
$x+y=a-----(1)$
and $ax-y=1------(2)$
Adding $(1)+(2)$
$\Rightarrow x(1+a)=1+a$
$\Rightarrow x=1$
Hence, $y=a-1$
Since it is in first quadrant,$a-1\ge 0$
$\Rightarrow a\ge 1$

Case $2:a<0$
$x+y=-a$ and $ax-y=1$
Solving For $x,y$
$x=\dfrac{1-a}{1+a}>0$
$\Rightarrow \dfrac{a-1}{a+1}<0$
$\Rightarrow a\epsilon (-1,1)$
also ,$y=-a-\left(  \dfrac{1-a}{1+a}\right )$ which should be $>0$
$\Rightarrow -\dfrac{a^2+1}{a+1}>0$
$\Rightarrow a<-1$
Combining both two cases, we get :
$a\ge 1$
$a\epsilon[1,\infty)$

Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

The line $y =\sqrt{2}x + 4\sqrt{2}$ is a normal to $y^{2} =4ax$ then a = 

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

The general form of equation of normal to the parabola $y^2=4ax$ is $y=mx-2am-am^3$.....(1) where $m$ is the slope of the normal.

According to the problem, $y =\sqrt{2}x + 4\sqrt{2}$.....(2) is the normal to the parabola.
Equation (1) and (2) are identical.
Then $m=\sqrt{2}$ and $-2am-am^3=4\sqrt{2}$ or, $a(-2\sqrt{2}-2\sqrt{2})=4\sqrt{2}$ or, $a=-1$ [ Using the value of $m$].
So $a=-1$.

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

The straight lines
$\left.\begin{matrix}
2kx-2y+3=0\
x+ky+2=0\
2x+k=0
\end{matrix}\right}k\in R$  pass through the same point for

  1. no real value of $k$
  2. exactly one real value of $k$
  3. three real values of $k$
  4. all real values of $k$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given equation of lines passes through same point i.e. lines are concurrent
$kx-2y+3=0\ 
x+ky+2=0\ 
2x+k=0$
$\Rightarrow \left| \begin{matrix} 2k & -2 & 3 \ 1 & k & 2 \ 2 & 0 & k \end{matrix} \right| =0$
$\Rightarrow k^{3}-2k-4=0$
$\Rightarrow (k-2)(k^2+2k+2)=0$
The discriminant of the quadratic expression is negative.
Hence there is only one real value of $k$

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

The three distinct straight lines $ax+by+c=0$;$bx+cy+a=0$ and $cx+ay+b=0$ are concurrent then

  1. $a+b+c=0$
  2. $a^{3}+b^{3}+c^{3}=3 abc$
  3. $a=b=c$
  4. $a^{2}+b^{2}+c^{2}=ab+bc+ca$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

The lines are concurrent,
$=>\begin{bmatrix}a&b& c\ b &c &a\ c & a& b\end{bmatrix}=0$
Applying,$ C _{1}=C _{1}+C _{2}+C _{3}$,
$=\begin{bmatrix}a+b+c&b& c\ a+b+c &c &a\ a+b+c & a& b\end{bmatrix}$
Applying, $R _{1}=R _{1}-R _{2}$ and $R _{2}=R _{2}-R _{3}$,
$=(a+b+c)\begin{bmatrix}0&b-c& c-a\ 0 &c-a &a-b\ 1 & a& b\end{bmatrix}$
Expanding by $C _{1}$,
$=(a+b+c)(-a^{2}-b^{2}-c^{2}+ab+bc+ac)=0$
$=> (a+b+c)=0$
or,
$(a+b+c)(-a^{2}-b^{2}-c^{2}+ab+bc+ac)=0$
$abc-a^{3}-b^{3}+abc+abc-c^{3}=0$
$a^{3}+b^{3}+c^{3}=3abc$
So options are A and B.

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

Find the equation of a line whose inclination is $\displaystyle 30^{\circ}$ and making an intercept of -3/5 on the y-axis

  1. $ y = \dfrac {5}{\sqrt{6}}x +\dfrac {2}{5} $
  2. $ y = \dfrac {1}{\sqrt{3}}x -\dfrac {3}{5} $
  3. $ y = \dfrac {3}{\sqrt{7}}x -\dfrac {1}{3} $
  4. $\displaystyle \dfrac{-5}{3}x$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation of any straight line can be written as $ y = mx + c $, where $m$ is its slope and $c$ is its y - intercept.

As inclination is $ 30^o $, slope of the line $ =  tan (30 ^o) = \dfrac {1}{\sqrt{3}} $

So equation of line is $ y = \dfrac {1}{\sqrt{3}}x  -\dfrac {3}{5} $