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

Joint equation of perpendicular lines passing through $(0,0)$ one of which is parallel to $6x-4y+3=0$ is

  1. $6x^{2}-5xy-6y^{2}=0$
  2. $6x^{2}+5xy-6y^{2}=0$
  3. $5x^{2}+5xy-6y^{2}=0$
  4. $6x^{2}-5xy-5y^{2}=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The line parallel to 6x - 4y + 3 = 0 can be written as 3x - 2y = 0. Its perpendicular line passing through the origin has the slope negative reciprocal, giving 2x + 3y = 0. The joint equation of the line and its perpendicular is found by multiplying their linear equations, resulting in (3x - 2y)(2x + 3y) = 6x^2 + 5xy - 6y^2 = 0.

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,
$\dfrac{x}{3}+\dfrac{y}{9}=1 \implies 3x+y=9$

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
Find the equation of the straight line whose $x$ and $y$-intercepts on the axes are given by $2$ and $3$.
  1. $3x-2y+6=0$
  2. $3x+2y-6=0$
  3. $3x-2y-6=0$

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

Consider the given points.

$(2, 0)$ and $(0, 3)$

We know that the equation of the line which is passing through the points
$y-y _1=\dfrac{y _2-y _1}{x _2-x _1}(x-x _1)$

So,

$y-0=\dfrac{3-0}{0-2}(x-2)$

$y=\dfrac{3}{-2}(x-2)$

$-2y=3x-6$

$3x+2y-6=0$

Hence, Option $B$ is the answer.

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 line passing through $(5,4)$ and $(-9,3)$

  1. $x-14y+51=0 $
  2. $ x+y=2 $
  3. $ x-2 y=0 $
  4. $ 2 x-y=0 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The equation of line Passing through $(5,4)$ and $(-9,3)$

The slope of line is $\dfrac{3-4}{5+9}=\dfrac 1{14}$

The equation of line is $y-4=\dfrac 1{14}(x-5)$

$\Rightarrow 14y-56=x-5$

$\Rightarrow x-14y+51=0$
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

Find the equation of a line which makes equal angles with the lines $x+y-2=0$ and $7x-y+4=0$ and passes through $(1, 2)$.

  1. $x-3y=5$
  2. $3x+y=5$
  3. $7x-y=5$
  4. $x+3y=7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The angle bisectors of the lines x+y-2=0 and 7x-y+4=0 are found using the formula (x+y-2)/sqrt(2) = +/- (7x-y+4)/sqrt(50). Simplifying these and checking which passes through (1, 2) gives the equation.

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

If $m$ and $b$ are real numbers and $mb > 0$, then the line whose equation is $y = mx + b$ cannot contain the point-

  1. $(0, 2009)$
  2. $(2009, 0)$
  3. $(0, -2009)$
  4. $(20, -100)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$y= mx + b$
for (2009,0)
substituting in the given line
we get $2009m+b=0$
that is possible only if $mb < 0$
which contradicts our initial assumption mb > 0
so option is $b$
Multiple choice maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

Consider the lines $2x+3y=0$,    $5x+4y=7$. Find the intersection point.

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

The lines are $2x+3y=0$......(1)


$x=\dfrac{-3y}{2}$

$5x+4y=7$........(2)

$5\left(\dfrac{-3y}{2}\right)+4y=7$

$-15y+8y=14$

$-7y=14$

$y=-2$

$x=\dfrac{-3(-2)}{2}=3$

$(x,y)=(3,-2)$

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

$y=2x+3$
Which of the following statements is true about the given line?

  1. The line passes through $(0,3)$ and $m=-2$
  2. The line passes through $(3,0)$ and $m=-2$
  3. The line passes through $(0,3)$ and $m=2$
  4. The line passes through $(3,0)$ and $m=2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Comparing the given equation $y=2x+3$ with $y = mx+c$, we get

Hence, $m=2$ and $c=3$.
So, options A and B are incorrect.

Option C:
Substitute $(0,3)$ in the given equation, we get
RHS: $=2(0)+3 = 3$
LHS: $y=3$
$LHS =  RHS$, Hence option C is correct.

Option D:
Substitute $(3,0)$ in the given equation, we get
RHS: $=2(3)+3 = 9$
LHS: $y=0$

$LHS \neq RHS$. Hence, option D is incorrect.

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 :$\displaystyle \frac{x-1}{3}-\frac{y+2}{2}=0$

  1. The line passes through $(4,0)$ and $m=2/3$
  2. The line passes through $(4,0)$ and $m=-2/3$
  3. The line passes through $(4,0)$ and $m=3/2$
  4. The line passes through $(4,0)$ and $m=-3/2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given line 
$\dfrac{x-1}{3}-\dfrac{y+2}{2}=0$

$\dfrac{x-1}{3}=\dfrac{y+2}{2}$

$2(x-1)=3(y+2)$

$2x-2=3y+6$

$y=\dfrac{2x}{3}-\dfrac{8}{3}$

on comparing above eq with $y=mx+c$

$slope(m)=\dfrac{2}{3}$

y-intercept$=-\dfrac{8}{3}$

when $y=0,x=4$

Hence it passes through $(4,0)$ with $m=\dfrac{2}{3}$