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

The equation $\displaystyle ax^{3}-9yx^{2}-y^{2}x+4y^{3}=0 $ represents three straight lines. If two of the lines are perpendicular to each other, then the value of $a$ is:

  1. 5

  2. -5

  3. 4

  4. -4
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

The given equation is $ax^3 - 9yx^2 -y^2x + 4y^3 = 0$, which represents three straight lines.


We can see that all the lines passes from origin $(0,0)$

Let's assume the lines are given by equation $y = mx$

Putting $y = mx$ in the given equation of three straight lines, we get,

$\Rightarrow ax^3- 9(mx)(x^2) - (mx)^2x + 4(mx)^3 = 0$

$\Rightarrow (a - 9m - m^2 + 4m^3) x^3 = 0$

$\Rightarrow 4m^3 -m^2 -9m +a = 0$ ....$(1)$

This equation in $m$ has three roots, $m _1$, $m _2$ and $m _3$, which are three slopes of three lines respectively.

If the two lines are perpendicular then let's assume $m _1.m _2 = -1$,

Product of roots in equation $(1)$ is $m _1.m _2.m _3 = \dfrac{-a}{4}$

Hence $m _3 = \dfrac{a}{4}$

also from equation $(1)$, $m _1 + m _2 + m _3 = \dfrac{1}{4}$

$\Rightarrow m _1m _2 + m _3(m _1 + m _2) = \dfrac{-9}{4}$

$\Rightarrow m _1 + m _2 = \dfrac{1-a}{4}$

$\Rightarrow a(1-a) = -20$

By Solving the above equation we get $a = 5, -4$

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

Equation $\displaystyle ax^{3}-9yx^{2}-y^{2}x+4y^{3}=0$ represents three straight lines. If two of the lines are perpendicular to each other then the value of a is

  1. 5

  2. -5

  3. 4

  4. -4
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

The given equation is $ax^3 - 9yx^2 -y^2x + 4y^3 = 0$, which represents three straight lines.


We can see that all the lines passes from origin $(0,0)$

Let's assume the lines are given by equation $y = mx$

Putting $y = mx$ in the given equation of three straight lines, we get,

$\Rightarrow ax^3- 9(mx)(x^2) - (mx)^2x + 4(mx)^3 = 0$

$\Rightarrow (a - 9m - m^2 + 4m^3) x^3 = 0$

$\Rightarrow 4m^3 -m^2 -9m +a = 0$ ....$(1)$

This equation in $m$ has three roots, $m _1$, $m _2$ and $m _3$, which are three slopes of three lines respectively.

If the two lines are perpendicular then let's assume $m _1.m _2 = -1$,

Product of roots in equation $(1)$ is $m _1.m _2.m _3 = \dfrac{-a}{4}$

Hence $m _3 = \dfrac{a}{4}$

also from equation $(1)$, $m _1 + m _2 + m _3 = \dfrac{1}{4}$

$\Rightarrow m _1m _2 + m _3(m _1 + m _2) = \dfrac{-9}{4}$

$\Rightarrow m _1 + m _2 = \dfrac{1-a}{4}$

$\Rightarrow a(1-a) = -20$

By Solving the above equation we get $a = 5, -4$

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

The pair of lines represented by $3ax^{2}+5xy+\left ( a^{2}-2 \right )y^{2}= 0$ and $\perp $ to each other for

  1. two values of $a$
  2. for all $a$
  3. for one value of $a$
  4. for no values of $a$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

 Using fact: Pair of lines $\displaystyle Ax^{2}+2hxy+By^{2}=0$ are 


$\displaystyle \perp $ to each other if $\displaystyle A+B=0$ 

$\displaystyle \Rightarrow 3a+a^{2}-2=0 $ $\displaystyle \Rightarrow a^{2}+3a-2=0 $ $\displaystyle\Rightarrow $ There exist two value of a as $\displaystyle D> 0$

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

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

  1. all real values of $d$
  2. two real values of $d$
  3. three real values of $d$
  4. no real value of $d$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $d$

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

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

  1. two values of $a$
  2. for all values of $a$
  3. for one value of $a$
  4. for no value of $a$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $a=\dfrac{-3-\sqrt{17}}{2}$

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

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

  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. $\dfrac{x-2}{2}=\dfrac{y-1}{-1}=\dfrac{z-1}{1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line must pass through the intersection of the given line and plane, and be perpendicular to the given line. Solving the system of equations for the intersection point and applying the cross product of the normal vector of the plane and the direction vector of the line yields the direction ratios.

Multiple choice maths lines equations of lines parallel to the x-axis and y-axis graphs of linear equations graph of linear equations in two variables

A straight line parallel to the $x$-axis has equation 

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

Let the equation of line be $y=mx+c  ...(1)$
Since, the line is parallel to $x$-axis, so slope of line, $m =0$
So, equation $(1)$ becomes $y=c$
Let the line is at a distance of $a$ from $x$-axis.
$a=c$
$\Rightarrow y=a$

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

The equation $a^2 x^2 + 2h(a+b) xy + b^2 y^2 = 0$ and $ax^2 + 2hxy + by^2 = 0$ represent

  1. two pairs of perpendicular straight lines

  2. two pairs of parallel straight lines

  3. two pairs of straight lines which are equally inclined to each other

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

$ax^2+2hxy+by^2=0$
Equation of the angle bisectors is given by
$\dfrac{x^2-y^2}{a-b}=\dfrac{xy}{h}$ ...(i)

For $ a^2x^2+2h(a+b)xy+b^2y^2=0$
The equations of the angle bisector is given by
$\dfrac{x^2-y^2}{a^2-b^2}=\dfrac{xy}{h(a+b)}$
 $\dfrac{x^2-y^2}{a-b}=\dfrac{xy}{h}$ ...(ii)
Since (i) is equal to (ii), the above lines are equally inclined to each other.

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

Which of the following does not represent a straight line?

  1. $ax+by+cz+d=0,ax+b'y+cz+d=0(b\neq b')$
  2. $ax+by+cz+d=0,a'x+by+cz+d=0(a\neq a')$
  3. $ax+by+cz+d=0,ax+by+cz+d'=0(d\neq d')$
  4. $ax+by+cz+d=0,ax+by+c'z+d=0(c\neq c')$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A. $ax+by+cz+d=0,ax+b'y+cz+d=0(b\neq b')$
Both the planes are different and are not parallel so they will definitely intersect on a line. Thus option A represents a line.
Similarly B and D represents a line. 
But C does not represents line. since $ax+by+cz+d=0,ax+by+cz+d'=0(d\neq d')$ represents two parallel planes which never intersects. 
Hence, option 'C' is correct choice.

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

Family of lines represented by the equation $(\cos \theta)x+(\cos \theta -\sin \theta)y-3(3\cos \theta+\sin \theta)=0$ passes through a fixed point $M$ for all real value of $\theta$. Find $M$ 

  1. $(6,3)$
  2. $(3,6)$
  3. $(-6,2)$
  4. $(3,-6)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let us consider the problem:

$\left( {\left( {\cos \theta  + \sin \theta } \right)x + \cos \theta  - \sin \theta } \right)y - 3\left( {3\cos \theta  + \sin \theta } \right) = 0$
$ \Rightarrow \cos \theta \left( {x + y - 9} \right) + \sin \theta \left( {x - y - 3} \right) = 0$
$ \Rightarrow $ $\left( {x + y - 9} \right) + \tan \theta \left( {x - y - 3} \right) = 0$
${L _1} + K{L _2} = 0$(pass through intersection of ${L _1}$ and ${L _2}$ for all value of $K$)
$x+y-9=0$
$ \Rightarrow $ $x - y - 3 = 0$ 
Hence,
$x+y=9$
$x-y=3$
hence the intersection point is $(6,3)$

Multiple choice mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

The equation of a line which passes through (2,3) and the product of whose intercepts on the coordinate axis is 27, can be

  1. 5x+4y=22

  2. 3x-y=3

  3. 3x+4y=18

  4. 2x+3y=13
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let the equation of the line is $\dfrac{x}{a}+\dfrac{y}{b}=1$
$\dfrac{2}{a}+\dfrac{3}{b}=1$

$2b+3a=ab$
$3a+2b=27$
$ab=27$
$b=\dfrac{27}{a}$
$3a+2\times \dfrac{27}{a}=27$
$3a^2+54=27a$
$3a^2-27a+54=0$
$a^2-9a+18=0$
$(a-6)(a-3)=0$
$a=6,3$
$b=\dfrac{9}{2}$ or $9$
Required equation is
$\dfrac{x}{6}+\dfrac{y}{\dfrac{9}{2}}=1 \implies 3x+4y=18$
Or,
$\dfrac{x}{3}+\dfrac{y}{9}=1 \implies 3x+y=9$