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 position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The image of the line $x-y-1=0$ in the line $2x-3y+1=0$ is

  1. $7x-17y+23=0$
  2. $17x-7y+23=0$
  3. $7x+17y+23=0$
  4. $ 17x+7y+23=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To find the image of a line in another line, find the intersection point and reflect a point from the first line. The intersection of x-y-1=0 and 2x-3y+1=0 is (2, 1). Reflecting a point like (1, 0) from the first line across the second line gives the new line equation.

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The equation of the line AB is y = x. if And B lie on the same side of the line mirror 2x - y = 1, then the equation of the image of AB is   _____________.

  1. x + y - 2 = 0

  2. 8x + y - 9 = 0

  3. 7x - y - 6 = 0

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

Reflecting the line y=x across 2x-y=1 involves reflecting two points on the line (e.g., (0,0) and (1,1)) across the mirror line and finding the equation of the line passing through the images.

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The equation of image of pair of lines $y=|x-1|$ with respect to y-axis is 

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

We have $y=|x-1|$


$\Rightarrow y^2=(x-1)^2$

Change $x$ by $-x$, then the required image is 

$y^2=(-x-1)^2$

$\Rightarrow y^2=x^2+2x+1$

$\Rightarrow x^2-y^2+2x+1=0$

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

The equation of the line which passes through $(0,0)$ and $(1,1)$ is ____________

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

The equation of the line which passes through (0,0) and (1,1) is y=x

As we know, the two points are $(0,0) (1,1)$
slope $ m$ $=\dfrac { { y } _{ 2 }-{ y } _{ 1 } }{ { x } _{ 2 }-{ x } _{ 1 } } $  $=\dfrac { 1-0 }{ 1-0 } =1$
Standard equation of the line is $y-{ y } _{ 1 }=m\left( x-{ x } _{ 1 } \right) $
Substituting value of $m =1 $ and point $(x,y) =  (0,0)$ we get:
$y-0=1\left( x-0 \right) $
 $y=x$ is the required equation of the line

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

The equation of a straight line passing through points $(0,0)$ and $(1,5)$ is given by:

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

Given a line passes through $A(0,0)$ & $B(1,5)$ 

We know that the equation of line which passes through $(a _1,b _1)$ & $\left( { a } _{ 2 },{ b } _{ 2 } \right) $ is 
$y-{ b } _{ 1 }=\cfrac { { b } _{ 2 }-{ b } _{ 1 } }{ { a } _{ 2 }-{ a } _{ 1 } } (x-{ a } _{ 1 })\ \therefore \quad y-0=\cfrac { 5-0 }{ 1-0 } (x-0)\ y=5x$ 

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

For any real value of $\lambda$, the equation $2x^2+3y^2-8x-6y+11-\lambda =0$ doesn't represents a pair of straight lines?

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The second degree terms cannot be factorized into two linear factors or else $h^2-ab=0-6=-$ve. Hence the given equation does not represent a pair of lines whatever $\lambda$ may be.

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

Let a and b be non-zero real numbers. Then, the equation $(ax^2+by^2+x)(x^2-5xy+6y^2)=0$ represents.

  1. Four straight lines, when $c=0$ and a, b are of the same sign
  2. Two straight lines and a circle, when $a=b$, and c is of sign opposite to that of a

  3. Two straight lines and a hyperbola, when a and b are of the same sign and c is of sign opposite to that of a

  4. A circle and an ellipse, when a and b are of the same sign and c is of a sign opposite to that of a
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation (ax^2 + by^2 + x)(x^2 - 5xy + 6y^2) = 0 represents a conic and two lines. If a = b, the first part represents a circle. If c (the constant term in the conic) is of opposite sign to a, it confirms the conic properties.

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

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

  1. A pair of lines

  2. Pair of lines and a circle

  3. A pair of lines and a parabola

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

Collecting the terms $y^2$ and y the given equation can be written as
$y^2(x^2-2x-3)-4y(x^2-2x-3)=0$
or $(x-3)(x+1)y(y-4)=0$
It represents four lines $x=-1$, $x=3$, $y=0$ and $y=4$.
These two sets of parallel lines form a square of side four.

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

Equation $4x^{2}+4xy-y^{2}-6x-3y-4=0$ represents a pair of parallel lines, then distance between these lines is

  1. $2\sqrt{5}$
  2. $\sqrt{5}$
  3. $\dfrac{2}{\sqrt{5}}$
  4. $\dfrac{3}{\sqrt{5}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the equation 4x^2 + 4xy - y^2 - 6x - 3y - 4 = 0, the distance between parallel lines is calculated using the formula 2 * sqrt(g^2 - ac) / sqrt(a(a+b)). Applying this to the given coefficients yields 2 * sqrt(5).

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

The four straight lines given by the equations $12x^2+7xy-12y^2=0$ and $12x^2+7xy-12y^2-x+7y-1=0$ lie along the sides of a 

  1. Square

  2. Rhombus

  3. Rectangle

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

taking $y$ is constant and finding the value of$x$ by roots formula.


$12x^2+(7y)x-12y^2=0$

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}=\dfrac{(7y)-\sqrt{(7y)^2-4(-12)y^212}}{2\times 12}$

$\dfrac{\Rightarrow x=-7y\pm\sqrt{49y^2+576y^2}{}}{2\times 12}$

$\Rightarrow 24x=-7y\pm \sqrt{625}y$

$\Rightarrow 24xy=-7y\pm 25y$.........(1)

Again $12x^2+(7y-1)x+(-12y^2-1+7y)=0$

$\therefore x=\dfrac{-(7y-1)\pm\sqrt{(7y-1)^2-4}(12)(7y-1-12y^2)}{24}$

$\Rightarrow x=\dfrac{1-7y\pm\sqrt{49y^2+1-14y-336y+48+576y^2}}{24}$

$\Rightarrow 24x=1-7y\pm\sqrt{625y^2-350y+49}$

$\Rightarrow 24x=1-7y\pm (25y-7)$

$\Rightarrow x=\dfrac{1-7y\pm 25y-7}{24}$........(ii)

from (i) and (ii) we can clearly see co.efficient of $x$ and $y$ are same so slope are sample $m _1=\dfrac{24}{18},m _2=-\dfrac{24}{32}$

So $m _1m _2= -1$

$\therefore $ it is a square

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

$3x^2+8xy-3y^2=0$ represents a pair of lines AB and BC, whereas the equation $3x^2+8xy-3y^2+2x=4y-1=0$ represents two lines CD and DA.
Answer the given question.
The equation of the CD is,

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

The equation 3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0 represents a pair of lines. By factoring this quadratic, we can identify the individual linear equations for the sides CD and DA.

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

The value of $k$ so that the equation $12{x}^{2}-10{y}^{2}+11x-5y+k=0$ may represent a pair of straight lines is

  1. $k=\dfrac{91}{48}$
  2. $k=\dfrac{94}{43}$
  3. $k=\dfrac{83}{23}$

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

$\Rightarrow$  The given equation is $12x^2-10y^2+11x-5y+k=0$

$\Rightarrow$  Comparing it with $ax^2+2hxy+by^2+2gx+2fy+c=0$
$\Rightarrow$  $a=12,\,b=-10,\,h=o,\,g=\dfrac{11}{2},\,f=\dfrac{-5}{2},\,c=k$
$\Rightarrow$  The condition is $abc+2fgh-af^2-bg^2-ch^2=0$
$\Rightarrow$  $12\times (-10)\times k+2\times (\dfrac{-5}{2})\times\dfrac{11}{2}\times 0-12\times (\dfrac{-5}{2})^2-(-10)\times (\dfrac{11}{2})^2-k\times (0)^2=0$
$\Rightarrow$  $-120k+0-75+\dfrac{605}{2}-0=0$
$\Rightarrow$  $\dfrac{-240k-150+605}{2}=0$
$\Rightarrow$  $-240k+455=0$
$\Rightarrow$  $k=\dfrac{455}{240}$
$\therefore$   $k=\dfrac{91}{48}$

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

If the pair of lines ${x^2}\, + \,2xy\, + \,a{y^2}\, = \,0$ and $a{x^2}\, + \,2xy\, + \,{y^2}\, = \,0$ have exactly one line in common, then joint equation of the other two lines is given by

  1. $3{x^2}\, + \,8xy\, - 3\,{y^2}\, = \,0$
  2. $3{x^2}\, + \,10xy\, + 3\,{y^2}\, = \,0$
  3. ${y^2}\, + \,2xy\, - 3\,{x^2}\, = \,0$
  4. ${x^2}\, + \,2xy\, - 3\,{y^2}\, = \,0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given the two pairs of lines, we find the common line by solving the intersection of the quadratic forms. The joint equation of the remaining two lines is found by dividing the combined equation by the common factor.