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

The equation of pair of lines joining origin to the points of intersection of $x^{2}+y^{2}=9$ and $x+y=3$ is

  1. $x^{2}+(3-x)^{2}=9$
  2. $xy=0$
  3. $(3+y)^{2}+y^{2}=9$
  4. $(x-y)^{2}=9$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Homogenizing the equation of the circle x^2 + y^2 = 9 using the linear equation x + y = 3 (which can be written as (x+y)/3 = 1) gives the combined equation of lines joining the origin to the intersection points. Simplifying this algebraic process leads to the result xy = 0.

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

If the pair of lines ${ ax }^{ 2 }+2hxy+{ by }^{ 2 }+2gx+2fy+c=0$ intersect on the y-axis, then

  1. $2fgh={ bg }^{ 2 }+{ ch }^{ 2 }$
  2. ${ bg }^{ 2 }\neq { ch }^{ 2 }$
  3. $abc=2fgh$
  4. $2fgh=af+{ ch }^{ 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the general second-degree equation to represent lines intersecting on the y-axis, the intersection point (0, y0) must satisfy the equation. Setting x=0 gives by^2 + 2fy + c = 0. The condition for the lines to intersect on the y-axis is derived from the general condition for intersection and the specific coordinate constraints.

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

The equation ${ x }^{ 2 }{ y }^{ 2 }-2x{ y }^{ 2 }-3{ y }^{ 2 }-4{ x }^{ 2 }y+8xy+12y=0$ represents 

  1. a pair of straight lines

  2. a pair of straight lines and a circle

  3. a pair of straight lines and a parabola

  4. a set of four lines forming a square
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given equation is 

${ x }^{ 2 }{ y }^{ 2 }-{ 2xy }^{ 2 }-{ 3y }^{ 2 }-{ 4x }^{ 2 }y+8xy+12y=0$
$\Rightarrow { y }^{ 2 }\left( { x }^{ 2 }-2x-3 \right) -4y\left( { x }^{ 2 }-2x-3 \right) =0$
$\Rightarrow { y }\left( { y }-4 \right) \left( { x }-3 \right) \left( x+1 \right) =0$
$\Rightarrow y=0,y=4,x=3,x=-1$
Hence, the equation represents four straight lines which evidently form a square.

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

Find the equation of a line which is perpendicular to the line joining $(4,2)$ and $(3,5)$ and cuts off an intercept of length $3$ units on $y$ axis. 

  1. $x-3y+9=0$
  2. $3x-y+6=0$
  3. $x-y+3=0$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let $(x _1,y _1)=(4,2),(x _2,y _2)=(3,5)$

Slope$(m)=\dfrac{{y} _{2}-{y} _{1}}{{x} _{2}-{x} _{1}}=\dfrac{5-2}{3-1}=-3$

Slope of a line perpendicular to the line $y=mx+c$ is $\dfrac{-1}{m}$ with $y-$intercept $3$ units.

$\therefore$ Slope of a line perpendicular to the line $y=mx+c$ is $\dfrac{-1}{-3}=\dfrac{1}{3}$ with $y-$intercept $c=3$ units.

Thus, the required equation is $y=\dfrac{1}{3}x+3$ or $3y=x+9$ or $x-3y+9=0$

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

The four sides of a quadrilateral are given by equ. $(xy+12-4x-4y{ ) }^{ 2 }=(2x-2y{ ) }^{ 2 }$. The equation of a line with slope $\sqrt { 3 } $ which divides the area of the quadrilateral in two equal parts is 

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

The equation (xy + 12 - 4x - 4y)^2 = (2x - 2y)^2 factors into (xy + 12 - 4x - 4y - 2x + 2y)(xy + 12 - 4x - 4y + 2x - 2y) = 0. This simplifies to (x-2)(y-6) = 0 and (x-6)(y-2) = 0, representing a rectangle with vertices (2,2), (6,2), (6,6), and (2,6). The center of this rectangle is (4,4), and any line passing through the center bisects the area.

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

If the equation  $2 x ^ { 2 } + 3 x y + b y ^ { 2 } - 11 x + 13 y + c = 0$  represents two perpendicular straight lines, then

  1. $b = - 2$
  2. $b = 2$
  3. $c = - 2$
  4. $c = 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a general equation ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 to represent perpendicular lines, the sum of the coefficients of x^2 and y^2 must be zero. Thus, a + b = 0. Given a = 2, we have 2 + b = 0, so b = -2.

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

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

  1. $2fgh= bg^{2}+ch^{2}$
  2. $bg^{2}\neq ch^{2}$
  3. $abc= 2fgh$

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

As $\displaystyle s=ax^{2}+2hxy +by^{2}+2gx +2fy+c=0 $ represent a pair of line $\displaystyle \therefore \begin{vmatrix}a &h  &g \h  &b  &f \g  &f  &c \end{vmatrix}=0$
or $\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0....(1)$ Now say point ofintersection onY axis be $\displaystyle (0,y _{1} $ and  point of intersection of pair of line be obtained by solving the equations $\displaystyle \frac{\partial s}{\partial x}=0=\frac{\partial s}{\partial y}$ $\displaystyle \therefore \frac{\partial s}{\partial x}=0\Rightarrow ax+by+g=0$ $\displaystyle \begin{matrix}\Rightarrow  \ \Rightarrow  \end{matrix} \left{\begin{matrix}hy _{1}+g=0 \by _{1}+f=0 \end{matrix}\right.>  ()$ and $\displaystyle \frac{\partial s}{\partial y}=0\Rightarrow bx+by+f=0$  On compairing the equation given in () we get $\displaystyle bg=fh $ and  $\displaystyle bg^{2}=fgh ....(2) $ Again $\displaystyle ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ meet at y-axis $\displaystyle \therefore x=0$ $\displaystyle \Rightarrow by^{2}+2fy+c=0$ whose roots must be equal $\displaystyle \Rightarrow by^{2}+2fy+c=0$ whose roots must be equal $\displaystyle \therefore f^{2}=bc af^{2}=abc ......(3)$ Now using (2) and (3) in equation (I) we have $\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$ $\displaystyle \Rightarrow (abc-af^{2})+(fgh-bg^{2})+fgh-ch^{2}=0$ $\displaystyle \Rightarrow 0+0+fgh-ch^{2}=0 \therefore ch^{2}=fgh .....(4) $ Now adding (2) and (4) $\displaystyle 2fgh=ch^{2}+bg^{2}$

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.