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 two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

If one of the lines given by $6x^{2}-xy+4cy^{2}= 0$ is $3x+4y= 0$, then $c$ equals

  1. 3

  2. -1

  3. 1

  4. -3
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation $\displaystyle ax^{2}+2hxy+by^{2}=0$

$\displaystyle  =(y-m _{1}x)(y-m _{2}x) $

$\displaystyle  \Rightarrow m _{1}+m _{2}=-\frac{2h}{b}=\frac{1}{4c }$.....(1) 

$\displaystyle  \Rightarrow m _{ _1}m _{2} =\frac{3}{2c} $
and $\displaystyle  3x+4y=0\Rightarrow m _{1} =-\dfrac {3}{4}$
 $\displaystyle \therefore m _{2} =-\frac{2}{c} $

 Now by $\displaystyle (1)$ we have $\displaystyle-\left(\frac{3}{4}+\frac{2}{c}\right) =\frac{1}{4c} $

$\displaystyle  \Rightarrow -\frac{3}{4}=\frac{1}{4c}+\frac{2}{c},\frac{3}{4}=\frac{1}{4c}+\frac{8}{4c}$ $\displaystyle  -\frac{3}{4}=\frac{9}{4c}$

$ \therefore c=-3$

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

If line $2x+7y-1=0$ intersect the lines $L _1=3x+4y+1=0$ and $L _2=6x+8y-3=0$ in $A$ and $B$ respectively, then equation of a line parallel to $L _1$ and $L _2$ and passes through a point $P$ such that $AP : PB=2:1$ (internally) is ($P$ is on the line $2x+7y-1=0$)

  1. $9x+12y+3=0$
  2. $9x+12y-3=0$
  3. $9x+12y-2=0$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Find points A and B by intersecting the line with L1 and L2. Use the section formula to find point P on the line 2x+7y-1=0 such that AP:PB=2:1. The required line is parallel to L1 and L2 (3x+4y+c=0) and passes through P.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

If the pair of lines represented by the equation $6x^{2}+17xy+12y^{2}+22x+31y+20=0$ be $2x+3y+p=0$ and $3x+4y+q=0$, then

  1. $p+q=9$
  2. $p^{2}+q^{2}=0$
  3. $3p+2q=22$
  4. $4p+3q=31$
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

$(2x+3y+p)(3x+4y+q)=0$

$6x^{2}+8xy+2qx+9xy+12y^{2}+3qy+3px+4py+pq=0$

$6x^{2}+17xy+12y^{2}+x(2q+3p)+y(3q+4p)+pq=0$
$\rightarrow 6x^{2}+17xy+12y^{2}+22x+31y+20=0$

Hence comparison gives us
$pq=20$
$3p+2q=22$
$4p+3q=31$
$(4p+3q)-(3p+2q)=31-22$
$p+q=9$.
Therefore
$p^{2}+q^{2}=(p+q)^{2}-2pq$
$=81-2(20)$
$=81-40$
$=41$

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

Joint equation of a pair of lines passing through the point of intersection of the lines $x^{2}+xy-2y^{2}-4x+7y-5=0$ and perpendicular to these lines is

  1. $2x^{2}-xy+y^{2}-4x+7y-5=0$
  2. $4x^{2}-7xy-2y^{2}+6x+15y-18=0$
  3. $2x^{2}+xy-y^{2}+2x+y+5=0$

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

$x^{ 2 }+xy-2y^{ 2 }-4x+7y-5=0\ \Rightarrow \left( x-y+1 \right) \left( x+2y-5 \right) =0$
Intersection point is $\left( 1,2 \right) $
Equation of line passing through $\left( 1,2 \right) $ and perpendicular to $x-y+1=0$ is
$x+y-3=0$
And equation of line passing through $\left( 1,2 \right) $ and perpendicular to $x+2y-5=0$ is
$2x-y=0$
Hence their joint equation is
$\left( x+y-3 \right) \left( 2x-y \right) =0$

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

Lines $x+y=4$, $3x+y=4$, $x+3y=4$ from a triangle which is

  1. Right-angled

  2. obtuse-angled isosceles

  3. acute-angled isosceles

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given

$x+y=4$........(1)
$3x+y=4$.......(2)
$x+3y=4$........(3)
solving (1) and (2)
$3x+y=4\ x+y=4\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ 2x=0$
$\therefore x=0\quad y=4\quad A=(0, 4)$
solving (1) and (3)
$x+3y=4\ x+y=4\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ 2y=0$
$\therefore y=0\quad x=4\quad B=(4, 0)$
solving (2) and (3)
$3x+y=4\ 3x+9y=12\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ -8y=-8$
$\therefore y=1\quad x=1\quad C=(1, 1)$
Distance between $A$ & $B$
$=\sqrt{(0-4)^2+(4-0)^2}$
$=\sqrt{4^2+4^2}$
$=\sqrt{32}=4\sqrt 2$
= Distance between $A$ & $C$
$\sqrt{(0-1)^2+(4-1)^2}$
$=\sqrt{1^2+3^2}$
$=\sqrt {10}$
Distance between $B$ & $C$
$\sqrt{(4-1)^2+(0-1)^2}$
$=\sqrt{10}$
As $AC=BC$
So, $\triangle ABC$, is isosceles triangle
using pythagoras theorem,
$AC^2+BC^2<AB^2$
as $10+10=20 < (4\sqrt 2)^2$
$20<32$
So $\triangle ABC$ is acute $\angle$  ed isosceles triangle.

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

If the pair of lines $a{ x }^{ 2 }+2hxy+b{ y }^{ 2 }+2gx+2fy+c=0$ intersect on the y axis then

  1. $2fgh=b{ g }^{ 2 }+c{ h }^{ 2 }$
  2. $b{ g }^{ 2 }\neq c{ h }^{ 2 }$
  3. $abc=2fgh$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given pairs of lines

$S=ax^2+2hxy+by^2+2gx+2fy+c=0------(1)$ intersects at y-axis $x=0$ 

Let coordinate of point of intersection is $(0,y)$
To find the point of intersection: 
$\dfrac{d{S}}{d{x}}=0$

$2ax+2hy+2g=0$

Here point of intersection is $(0,y)$ SO above eq passes through $(0,y)$
$0+2hy=-2g$

$y=-\dfrac{g}{h}$
So Intersection point $\left (0,-\dfrac{g}{h} \right)$

Given pair passes through $\left (0,-\dfrac{g}{h} \right)$

$b\left( -\dfrac{g}{h}\right)^2+2f\left ( -\dfrac{g}{h} \right )+c=0$

$bg^2-2fgh+ch^2=0$

$2fgh=bg^2+ch^2$

Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

Find the equation of the line passing through $(-3,5)$ and perpendicular to the line through the points $(2,5)$ and $(-3,6)$.

  1. $5x-2y+20=0$
  2. $x-5y+20=0$
  3. $5x-y+20=0$
  4. $5x+y+20=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The slope of the line, whose end points is $\left( {2,5} \right)$ and $\left( { - 3,6} \right)$ is,

$m = \frac{{6 - 5}}{{ - 3 - 2}}$

$ =  - \frac{1}{5}$

The two non-vertical lines are perpendicular to each other if they have the slopes as negative reciprocals of each other.

So, the slope of the line that is perpendicular to the line joining the points $\left( {2,5} \right)$ and $\left( { - 3,6} \right)$ is,

$ =  - \frac{1}{m}$

$ = 5$

The equation of line passing through the point $\left( { - 3,5} \right)$ with slope 5 is,

$\left( {y - 5} \right) = 5\left( {x - \left( { - 3} \right)} \right)$

$y - 5 = 5x + 15$

$5x - y + 20 = 0$

Therefore, the required equation of line is $5x - y + 20 = 0$.


Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

If the pair of lines $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ intersecty on y-axis then

  1. $2fgh = b{g^2} + c{h^2}$
  2. $b{g^2} \ne c{h^2}$
  3. $abc = 2fgh$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Put $x = 0$ in the given equation
$\Rightarrow\,b{y}^{2}+2fy+c=0$.
For unique point of intersection ${f}^{2}−bc=0$
$\Rightarrow\,a{f}^{2}−abc=0$.
Since $abc+2fgh−a{f}^{2}−b{g}^{2}−c{h}^{2}=0$
$\Rightarrow\,2fgh−b{g}^{2}−c{h}^{2}=0$
$\therefore\,2fgh=b{g}^{2}+c{h}^{2}$
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

Two pair of straight lines have the equation $\displaystyle x^{2}+6xy+9y^{2}=0: : and: : ax^{2}+2bxy+cy^{2}=0 $. If one line among them is common, then the value of $9a - 6b + c$ is

  1. $1$
  2. $3$
  3. $0$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given pairs 
$x^2+6xy+9y^2=0$

$(x+3y)^2=0\Rightarrow x=-3y$

On comparing $3y=-x$ with $y=mx+c$ we get slope 

$m=-\dfrac{1}{3}$
Second pair of line 

$cy^2+2bxy+ax^2=0$

$y=\dfrac{-2bx\pm\sqrt{4b^2x^2-4acx^2}}{2c}$

$y=\dfrac{-2bx\pm2x\sqrt{b^2-ac}}{2c}$


$y=\left (\dfrac{-b\pm \sqrt{b^2-ac}}{c}  \right )x$

On comparing above eq with $y=mx+c$ we get 

$m _{1}=\dfrac{-b+ \sqrt{b^2-ac}}{c}$ and $m _{2}=\dfrac{-b- \sqrt{b^2-ac}}{c}$

Here one line is common in both pairs so slope will be same 

$m _{1}=-\dfrac{1}{3}$

$\dfrac{-b+ \sqrt{b^2-ac}}{c}=-\dfrac{1}{3}$

$-3b+3 \sqrt{b^2-ac}=-c$

$3 \sqrt{b^2-ac}=-c+3b$

On squaring both sides 
$9b^2-9ac=c^2+9b^2-6bc$
$9ac+c^2-6bc=0$
$c(9a-6b+c)=0$
$c=0$ and $9a-6b+c=0$
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

If the equation $ax^{3}+3bx^{2}y+3cxy^{2}+dy^{3}=0$ $(a, b,c, d\neq 0)$ represents three coincident lines, then 

  1. $a=c$
  2. $b=d$
  3. $\displaystyle {\frac{a}{b}=\frac{b}{c}=\frac{c}{d}}$
  4. $ac=bd$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ax^3+3bx^2y+3cxy^2+dy^3=0$  ---(1)   represent three coincident line

Let $ y=mx$ is that line

$(y-mx)^3=0$

$y^3+3m^2xy-3y^2(mx)-m^3n^3=0$

$m^3x^3+3mny^2-3m^2xy-y^3=0$   ---(2)

Compare 1 & 2,

$\dfrac{a}{m^3}=\dfrac{b}{m}=\dfrac{c}{-m^2}=\dfrac{d}{-1}$

$\therefore \dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}$

Multiple choice maths constructions mid-point formula midpoints division of a line segment

What is the y intercept of the line that is parallel to $y=3x,$ and which bisects the area of rectangle with corners at $(0,0), (4,0) ,(4,2) $ and $(0,2)$? 

  1. $ -7$
  2. $-6$
  3. $ -5$
  4. $ -4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Rectangle's midpoints are $ = \left( {\frac{{0 + 4}}{2},\frac{{0 + 2}}{2}} \right) = \left( {\frac{4}{2},\frac{2}{2}} \right) = \left( {2,1} \right)$

Slope line y$=$ 3x will be :
${m _1} = 3$
Parallel line will be
${m _2} = 3$
Equation of line passing through (2 , 1)
$y - {y _1} = m(x - {x _1})$
$(y - 1) = 3(x - 2)$
$y - 1 = 3x - 6$
$y = 3x - 5$
Hence on comparing 
Y- intercept $=$ - 5

Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

IF the lines $ \displaystyle y=m _{1}x+c $  and $  y=m _{2}x+c _{2}  $ are parallel , then 

  1. $ \displaystyle m _{1}=m _{2} $
  2. $ \displaystyle m _{1}=m _{2} =1 $
  3. $ \displaystyle m _{1}=m _{2} =-1 $
  4. $ \displaystyle m _{1}=m _{2} =0 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Two lines are said to be parallel if the slopes of two line will be equal
$m _1=m _2$

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

If line $y+3x=c$ is normal of the ellipse ${ x }^{ 2 }+3{ y }^{ 2 }=3$ then equation of normal is-

  1. $y-3x\pm \sqrt { 3 } =0$
  2. $y+3x\pm \sqrt { 3 } =0$
  3. $y+3x\pm 3 =0$
  4. $y+3x\pm 1 =0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If $y=mx+c$ is normal to ellipse
${ c }^{ 2 }={ m }^{ 2 }\cfrac { { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 }{ m }^{ 2 } } $
${ x }^{ 2 }+3{ y }^{ 2 }=3$
$\cfrac { { x }^{ 2 } }{ 3 } +\cfrac { { y }^{ 2 } }{ 12 } =1$
${ a }^{ 2 }=3,b=1$
$y+3x=c$
$m=-3$
${ c }^{ 2 }={ (-3) }^{ 2 }\cfrac { { \left( { a }^{ 2 }-{ b }^{ 2 } \right)  }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 }{ m }^{ 2 } } =9\times \cfrac { { (3-1) }^{ 2 } }{ 3+9 } $
$=\cfrac { 9\times 4 }{ 12 } =3$

Equation of normal
$y+3x\pm \sqrt { 3 } =0$