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 functions and graphs different forms of equation of a line

The equation of a straight line passing through a point $(-5,4)$ and which cuts off an intercept of $\sqrt{2}$ units between the lines $x+y+1=0$ and $x+y-1=0$ is

  1. $x-2y-13=0$
  2. $2x-y+14=0$
  3. $x-y+9=0$
  4. $x-y+10=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the given point  be $A=(−5,4)$ and the given lines be $l _1 \rightarrow x+y+1=0$ and  $l _2\rightarrow x+y-1=0$

The  point $A$ lies on $l _1$


If segment $AM\perp l _2$ and $M$ lies on $l _2$, then, the distance $ AM$ is given by,


$\Rightarrow AM=\dfrac{|−5+4−1|}{\sqrt{1^2+1^2}}=\dfrac{2}{\sqrt 2}=\sqrt 2$


$\Rightarrow $ This means that if $B$ is any point  on $l _2$ then  $AB>AM$. No line other than $AM$ cuts off an intercept of

length $\sqrt 2$ between $l _1$ and $l _2$.


$\Rightarrow $To determine the equation of $AM$, we need to find the co-ordinates of the Point $M$


Since, $AM\perp l _2$ and  the slope $l _2$ is $−1$, the slope of$AM$ must be $1$.  Also  $A(−5,4)$ lies on $AM$


By the point slope formula, the equation of the required line is


$\Rightarrow  y−4=1(x−(−5))$

$\Rightarrow y-4=x+5$

$\Rightarrow x−y+9=0$

Multiple choice maths functions and graphs different forms of equation of a line

Equation of a straight line passing through the point $(4, 5)$ and equally inclined to the lines $3x=4y+7$ and $5y=12x+6$ is?

  1. $9x-7y=1$
  2. $9x+7y=71$
  3. $7x+9y=73$
  4. $7x-9y+17=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The slopes of the given lines are m₁ = 3/4 and m₂ = 12/5. For a line to be equally inclined to both, its slope m must satisfy |(m-m₁)/(1+mm₁)| = |(m-m₂)/(1+mm₂)|. Solving gives two possible slopes: m = -7/9 (internal bisector) or m = 9/7 (external bisector). Using point (4,5) with slope -7/9: y-5 = (-7/9)(x-4), giving 9y-45 = -7x+28, or 7x+9y=73.

Multiple choice maths functions and graphs different forms of equation of a line

The equation of a straight line passing through the point (-5,4) and which cuts off an intercept of $\sqrt { 2 } $ unit between the lines $x+y+1=0$ and $x+y-1=0$ is:

  1. $2x-y+14=0$
  2. $3x+y+11=0$
  3. $x-y+9=0$
  4. $4x+5y=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the given point be $A=(-5,4)$ and the given lines be,

$L _1:x+y+1=0$ and 
$L _2:x+y-1=0$
Observe that, $A\in L _1$.
If segment $AM\perp L _2,$ $M\in L _2,$ then, the distance $AM$ is given  by,

$\Rightarrow$  $AM=\dfrac{|-5+4-1|}{\sqrt{1^2+1^2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}$

This means that if $B$ is any point on $L _2,$ then, $AB>AM.$
In other words, no line other than $AM$ cuts off an intercept of length $\sqrt{2}$ between $L _1,$ and $L _2$ or $AM$ is the required line.
To determine the equation of $AM,$ we need to find the co-ordinates of the point $M$.
Since, $AM\perp L _2,$ and the slope of $L _2$ is $-1,$ the slope of $AM$ must be $1.$
Further, $A(-5,4)\in AM$
By the slope-point form the equation of the required line is,
$\Rightarrow$  $y-4=1(x-(-5))$
$\Rightarrow$  $y-4=x+5$
$\Rightarrow$  $x-y+9=0$

Multiple choice maths functions and graphs different forms of equation of a line

The equation of a straight line passing through the point (-5,  4) and which cuts off in intercept of $\sqrt { 2 } $ unit. between the lines $x+y+1=0$ and $x+y-1=0$ is:

  1. $2x-y+14=0$
  2. $3x+y+11=0$
  3. $x-y+9=0$
  4. $4x+5y=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the given point be $A\left( {{x} _{1}},{{y} _{1}} \right)=\left( -5,4 \right)$ and the given lines be

$ x+y+1=0\,\,......\,\,\left( 1 \right) $

$ x+y-1=0\,\,......\,\,\left( 2 \right) $

Observe that,

From equation (1) to,

Let AM is perpendicular distance

Then, $AM=\dfrac{\left| -5+4-1 \right|}{\sqrt{{{1}^{2}}+{{1}^{2}}}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}$

Let B is any point on equation $(2)$to,

From equation (2)

$ x+y-1=0 $

$ y=-x+1 $

On comparing that,

$y=mx+c$

Then, $m=-1$

Slope of perpendicular line is

${{m} _{1}}=\dfrac{-1}{m}=1$

Then, equation of line is

$ y-{{y} _{1}}=m\left( x-{{x} _{1}} \right) $

$ \Rightarrow y-4=1\left( x+5 \right) $

$ \Rightarrow y-4=x+5 $

$ \Rightarrow x-y+5+4=0 $

$ \Rightarrow x-y+9=0 $

Hence, this is the answer.

Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

If the lines $L _{1}:\lambda ^{2}x-y-1=0$ $L _{2}:x-\lambda ^{2}y+1=0$ $L _{3}:x+y-\lambda ^{2}=0$ pass through the same point the value(s) of $\lambda$ equals

  1. $1$
  2. $\sqrt{2}$
  3. $2$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The given equations passes through same point.So, they are concurrent lines

$\Rightarrow \left| \begin{matrix} { \lambda  }^{ 2 } & -1 & -1 \ 1 & -{ \lambda  }^{ 2 } & 1 \ 1 & 1 & -{ \lambda  }^{ 2 } \end{matrix} \right| =0$

$\Rightarrow { \lambda  }^{ 6 }-3{ \lambda  }^{ 2 }-2=0$

By doing synthetic division we get,
$(\lambda^4-2\lambda^2+1)(\lambda^2-2)=0$
$(\lambda^2-1)^2(\lambda^2-2)=0$
$\lambda=\pm \sqrt 2 or \lambda=\pm1$
But here $\lambda=\pm1\  does\  not\  satisfies\  ,hence\  \lambda=\sqrt2$
Option B satisfies above equation

Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

Number of values of $a$ for which the lines $2x+y-1=0, ax+3y-3=0, 3x+2y-2=0$ are concurrent is

  1. 0

  2. 1

  3. 2
  4. $\infty$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Here coefficient matrix,  $\Delta = \begin{vmatrix} 2 & 1 & -1 \ a & 3 & -3 \ 3 & 2 & -2 \end{vmatrix}$
Using $C _2 \to C _2+C _3$
$\Delta = \begin{vmatrix} 2 & 0 & -1 \ a & 0 & 3 \ 3 & 0 & -2 \end{vmatrix}=0$
Clearly $\Delta=0$, Hence given lines are concurrent for all values of $a$

Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

If the lines $2\mathrm{x}-\mathrm{a}\mathrm{y}+1 =0,\ 3\mathrm{x}-\mathrm{b}\mathrm{y}+1 =0,\ 4\mathrm{x}-\mathrm{c}\mathrm{y}+1 =0$ are concurrent then $\mathrm{a}$, b,c are in ?

.

  1. G.P.

  2. A.P.

  3. H.P.

  4. A.G.P.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$2x-ay+1 =0,\ 3x-by+1 =0,\ 4x-cy+1 =0$


Given lines are concurrent

$\Rightarrow \begin{vmatrix} 2 & -a & 1 \ 3 & -b & 1 \ 4 & -c & 1 \end{vmatrix}=0$

$\Rightarrow  2b-a-c=0$

$\Rightarrow 2b=a+c$

Hence, a,b,c are in A.P.

Multiple choice lines in planes applications of determinants inverse of a matrix and linear equations matrix algebra maths

The straight lines $\mathrm{x}+2\mathrm{y}-9=0,3\mathrm{x}+5\mathrm{y}-5=0$ and $\mathrm{a}\mathrm{x}+\mathrm{b}\mathrm{y}-1=0$ are concurrent if the straight line $22\mathrm{x}-35\mathrm{y}-1=0$ passes through the point 

  1. (a, b)

  2. (b,a)

  3. (-a,b)

  4. (-a, -b)
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$x+2y-9=0$ ---(1)


$3x+5y-5=0$ ---(2)


$ax+by-1=0$ ---(3)

Solving (1) and (2) simultaneously we get 

$y=22, x=-35$

Now, equation (1), (2) and (3) will be concurrent, that is they will pass through one point if $y=22, x=-35$ satisfy the third equation $ax+by-1=0$.

Substituting the values of 'x' and 'y' in this equation we get $-35a+22b-1=0$ ---(4)


And another equation given is $22x-35y-1=0$ ---(5)

Equation (5) will be of the form of equation (4), if we substitute

$x=b$ & $y=a$

That is $22x-35y-1=0$ passes through $(b,a)$

Multiple choice maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

Two lines $L _{1} :2x+3y-5=0$ and $L _{2} :3x-4y+1=0$ intersect a point $P$ and make an angle $\theta$ with each other. Equation of a line which passes through $P$ and makes an angle $(\pi/2-\theta)$ with the line $L _{1}$ is

  1. $16x+64y+79=0$ and $4x+3y+7=0$
  2. $16x+63y-79=0$ and $4x+3y+7=0$
  3. $16x-63y+79=0$ and $4x-3y+7=0$
  4. $16x-63y-79=0$ and $4x-3y+7=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice mathematics and statistics coordinates, points and lines general equation of a line: ax+by+c=0 general equation of a line reducing equation of straight line to standard form
If the line $(2x+y+1)+\lambda(x-y+1)=0$ is parallel to $y-axis$ then value of $\lambda$ is ?
  1. $1$
  2. $-1$
  3. $\dfrac{1}{2}$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Parallel to y-axis $\Rightarrow y=0$
$2x+y+1+\lambda x-\lambda y+\lambda =0$
$x(2+\lambda )+y(1-\lambda )+(1+\lambda )=0$
coefficient of $y=0$
$1-\lambda =0\Rightarrow \lambda =1$
Multiple choice mathematics and statistics coordinates, points and lines general equation of a line: ax+by+c=0 general equation of a line reducing equation of straight line to standard form

The equation of the line passing through $(-4, 3)$, parallel to the $3x+7y+6=0$

  1. 3x+7y-9=0

  2. 3x+7y+9=0

  3. 3x+7y+3=0

  4. 3x+7y+12=0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line parallel to $3x+7y+6=0$ is $3x+7y+k=0$

It passes through $(-4,3)$
$\implies 3(-4)+7(3)+k=0\-12+21+k=0\\implies k=-9$
So the required equation is $3x+7y-9=0$

Multiple choice mathematics and statistics coordinates, points and lines general equation of a line: ax+by+c=0 general equation of a line reducing equation of straight line to standard form

The slope and  the y-intercept  of the given line, $y-3x -6=0$ are respectively,

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

The given equation is $y-3x-6=0$ ........ $(1)$


To obtain the slope and $y-$intercept of the given equation, we write it in slope-intercept form which is 


$y=mx+c$, where $m$ and $c$ are slope and $y-$intercept

From $(1)$,

$y-3x-6=0\implies y=3x+6$

Comparing it with $y=mx+c$ we get,

slope$=m=3$ and y-intercept$=c=6$

Hence, option C is correct.

Multiple choice mathematics and statistics coordinates, points and lines general equation of a line: ax+by+c=0 general equation of a line reducing equation of straight line to standard form

The slope and y-intercept of the following line are respectively

$2y + 2x - 5 = 0$

  1. $ slope=m=1\quad and\quad y-intercept=c=\frac { 5 }{ 2 } . $
  2. $ slope=m=1/5\quad and\quad y-intercept=c=\frac { 2 }{ 5 } . $
  3. $ slope=m=-1\quad and\quad y-intercept=c=\frac { 5 }{ 2 } . $
  4. $ slope=m=-1/5\quad and\quad y-intercept=c=\frac { 2 }{ 5 } . $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given equation is $2y+2x-5=0$ ........ $(1)$

To obtain the slope and $y-$intercept of the given equation, we write it in slope-intercept form which is 
$y=mx+c$, where $m$ and $c$ are slope and $y-$intercept

From $(1)$,
$2y+2x-5=0\implies y=-x+\dfrac{5}{2}$
Comparing it with $y=mx+c$ we get,
slope$=m=-1$ and y-intercept$=c=\dfrac{5}{2}$
Hence, option C is correct.

Multiple choice mathematics and statistics coordinates, points and lines general equation of a line: ax+by+c=0 general equation of a line reducing equation of straight line to standard form

The slope and y-intercept of the following line are respectively

$7x-y + 3 =0$

  1. $ slope=m=7/3\quad and\quad y-intercept=1.\\ $
  2. $ slope=m=-7\quad and\quad y-intercept=3.\\ $
  3. $ slope=m=-7/3\quad and\quad y-intercept=1.\\ $
  4. $ slope=m=7\quad and\quad y-intercept=3.\\ $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The given equation is $7x-y+3=0$ ........ $(1)$

To obtain the slope and $y-$intercept of the given equation, we write it in slope-intercept form which is 
$y=mx+c$, where $m$ and $c$ are slope and $y-$intercept

From $(1)$,
$7x-y+3=0\implies y=7x+3$
Comparing it with $y=mx+c$ we get,
slope$=m=7$ and y-intercept$=c=3$
Hence, option D is correct.