Mathematics

Straight Lines and Coordinates

155 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

Consider the line: y= -x +4

Which of the following is correct.

  1. The line passes through (0,4) and m=1.

  2. The line passes through (0,4) and m=-1.

  3. The line passes through (0,0) and m=-1.

  4. The line passes through (4,0) and m=-1.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given line 
$y=-x+4$
on comparing above eq with $y=mx+c$
$slope(m)=-1$
y-intercept$=4$
Hence it passes through (0,4) with $m=-1$
Multiple choice maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

Consider the equation of the line $\displaystyle x-3=\frac{2}{5}\left ( y-1 \right )$. Which of the following is correct?

  1. The line passes through $(6,5)$ and $m=-2/5$.
  2. The line passes through $(5,6)$ and $m=-5/2$.
  3. The line passes through $(6,5)$ and $m=2/5$.
  4. The line passes through $(5,6)$ and $m=5/2$.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given line 
$x-3=\dfrac{2}{5}(y-1)$
$5(x-3)=2(y-1)$
$5x-15=2y-2$
$2y=5x-13$
$y=\dfrac{5x}{2}-\dfrac{13}{2}$
on comparing above eq with $y=mx+c$
$slope(m)=\dfrac{5}{2}$

when $x=5$
$2y=25-13$
$2y=12$
$y=6$
Hence it passes through (5,6) with $m=\dfrac{5}{2}$
Multiple choice maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

For the pair of linear equations given below, draw graph and then state, whether the lines drawn are,
$\displaystyle y=3x-1$
$\displaystyle \frac{x}{2}+\frac{y}{3}=1$

  1. Perpendicular

  2. Parallel

  3. Intersecting but not at right angles

  4. Options B & C
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The first line is y = 3x - 1 (slope m1 = 3). The second line x/2 + y/3 = 1 can be rewritten as y = -3/2x + 3 (slope m2 = -3/2). Since m1 is not equal to m2 and their product is not -1, the lines intersect but are not perpendicular.

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

For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are 
$\displaystyle 3x+4y=24$
$\displaystyle \frac{x}{4}+\frac{y}{3}=1$

  1. intersecting but not at right anglesl

  2. Options B & D

  3. perpendicular

  4. parallel
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The first line 3x + 4y = 24 has a slope of -3/4. The second line x/4 + y/3 = 1 can be rewritten as 3x + 4y = 12, which also has a slope of -3/4. Since the slopes are equal but the intercepts are different, the lines are parallel.

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

The straight lines given by the equations $\displaystyle x+y=2 , x-2y=5 \ and \ \frac{x}{3}+y=0$ are?

  1. concurrent

  2. intersecting to make a right triangle.

  3. intersecting to make an isosceles triangle.

  4. parallel to each other.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given lines
$x+y=2$------(1)
$x-2y=5$----(2) and $\dfrac{x}{3}+y=0$----(3)
Solving eq (1) and (2)
$x-2(2-x)=5$
$x-4+2x=5$
$x=3$ and $y=2-3=-1$
Point of intersection of line (1) and (2) is $P(3,-1)$
Solving eq (2) and (3)
$-3y-2y=5$
$-5y=5$
$y=-1$ and $x=-3y=3$
Point of intersection of line (2) and (3) is $Q(3,-1)$
Solving eq (1) and (3)
$-3y+y=2$
$-2y=2$
$y=-1$ and $x=-3y=3$
Point of intersection of line (1) and (3) is $R(3,-1)$
Here point of intersection of all line is same Hence line is concurrent
Multiple choice maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

If the line ax + by + c = 0 is such that  a = 0 and b, $\displaystyle c\neq 0$ then the line is perpendicular to 

  1. x-axis

  2. y-axis

  3. x + y =1

  4. x = y
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When $ a= 0 $ then the line equation becomes $ by + c = 0 $ or $ y = -\frac {c}{b} $

Equations of the form $ y =k $ are parallel to x-axis. This also means that they are perpendicular to $ y - $ axis as $ x-$ axis and $ y- $axis are perpendicular to each other.

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

A line passing through (2, 2) is perpendicular to the line $3x+y=3$. Its y intercept is _____________.

  1. $\dfrac { 1 }{ 3 } $
  2. $\dfrac { 2 }{ 3 } $

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

The given line 3x + y = 3 has a slope of -3. A perpendicular line must have the negative reciprocal slope, which is 1/3. Using the point-slope form with point (2, 2) gives y - 2 = (1/3)(x - 2), which simplifies to y = (1/3)x + 4/3, making the y-intercept equal to 4/3.

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 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 :
$a\ge 1$
$a\epsilon[1,\infty)$