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 physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

The line $y =\sqrt{2}x + 4\sqrt{2}$ is a normal to $y^{2} =4ax$ then a = 

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

The general form of equation of normal to the parabola $y^2=4ax$ is $y=mx-2am-am^3$.....(1) where $m$ is the slope of the normal.

According to the problem, $y =\sqrt{2}x + 4\sqrt{2}$.....(2) is the normal to the parabola.
Equation (1) and (2) are identical.
Then $m=\sqrt{2}$ and $-2am-am^3=4\sqrt{2}$ or, $a(-2\sqrt{2}-2\sqrt{2})=4\sqrt{2}$ or, $a=-1$ [ Using the value of $m$].
So $a=-1$.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The points with position vectors $60\hat{i}+3\hat{j}$, $40\hat{i}-8\hat{j}$, $a\hat{i}-52\hat{j}$  are collinear if

  1. $a=-40$
  2. $a=40$
  3. $a=20$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

suppose ${60i + 3j}$ , ${40i - 8j}$ and ${ai - 52j}$ is the three position of vector $A,B,C$


$\begin{array}{l} \overrightarrow { AB } =\left( { 40i-8j } \right) -\left( { 60i+3j } \right)  \ \overrightarrow { AB } =-20i-11j \ \overrightarrow { BC } =\left( { ai-52j } \right) -\left( { 40i-8j } \right)  \ \overrightarrow { BC } =\left( { a-40 } \right) i-44j \ \left( { a-40 } \right) i-44j=m\left( { -20i-11j } \right)  \ \left( { a-40 } \right) i-44j=-20im-11jm \ -44=-11m \ m=\frac { { -44 } }{ { -11 } }  \ m=4 \ a-40=-20m \ a-40=-20\left( 4 \right)  \ a=-80+40 \ a=-40 \end{array}$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The points with position vectors $ 60i + 3j,  40i -8j$ and $ ai -52j $ are collinear if

  1. $a = -40$
  2. $a = 40$
  3. $a = 20$

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

Denoting $a,b,c$ by the given vectors respectively
These vectors will be collinear if there is some constant $k$ such that $c-a=K\left( b-a \right) $
$\Rightarrow a-60=-20K$ and $-55=-11K$
$\Rightarrow a=-100+60=-40$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points $(\alpha, - 1), (2, 1)$ and $(4, 5)$ are collinear, then find $\alpha $ by vector method.

  1. $4$
  2. $1$
  3. $8$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
If there points are collinear then vectors from one to another will have scalar triple produced $0$.Point $\left(\alpha,-1\right), \left(2,1\right), \left(4,5\right)$
$\left( 2-\alpha  \right) \hat { i } +2\hat { j } -\bar { A }$
$2\hat { i } +4\hat { j } -\bar { B }$
$\left( 4-\alpha  \right) \hat { i } +6\hat { j } -\bar { C }$
$ \bar { A } .\left( \bar { B } \times \bar { C }  \right) =0$
$\left( \left( 2-\alpha  \right) \hat { i } +2\hat { j }  \right) \left( 2\hat { i } +4\hat { j }  \right) \times \left( \left( 4-\alpha  \right) \hat { i } +6\hat { j }  \right) \\ \left( \left( 2-\alpha  \right) \hat { i } +2\hat { j }  \right) .\left[ 12\hat { k } -16\hat { k } +4\alpha \hat { k }  \right] =0$
$4\alpha =4$
 $\alpha =1$
Also the direction vector will be proportion
$\left( 2-\alpha,2 \right)=\lambda\left( 4-2.5-1\right)$
$\left( 2-\alpha,2 \right)=\lambda\left( 2,4\right)$
$\lambda=\dfrac{1}{2}$ as $2=4\lambda$
$2-\alpha=1$
$\therefore \alpha=1$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points $a(1, 2, -1), B(2, 6, 2)$ and $c(\lambda, -2, -4)$ are collinear then $\lambda$ is

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

D.R of AB are $ 2 -1, 6-2,2-(-1) i.e. 1,4,3$

D.R. of AC are $ λ.−1,−2−2,−4−(−1) $
$i.e., λ−1,−4,−3 $
Since A, B, C are collinear $\therefore AB||BC $
$\therefore \dfrac {\lambda-1}{1} =\dfrac{-4}{4}=\dfrac{-3}{3}$
$\Longrightarrow \lambda-1=-1$
$\therefore \lambda=0$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points (p. 0), (0, q) and (1, 1) are collinear then $\dfrac { 1 }{ p } +\dfrac { 1 }{ q } $ is equal to 

  1. -1

  2. 1

  3. 2

  4. 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
If the area of triangle is zero, then the points are collinear.
Points are collinear.
Point are $(p, o)(o, q)(i,q)$

$\Delta =\dfrac {1}{2}[p(q-1)-0(1-0)+1(0-q)]$
$\Rightarrow \dfrac{1}{2}[p(q-1)-q]$
$\Rightarrow \dfrac{1}{2}[p(q-1)-q]$

$\Rightarrow \dfrac{1}{2}[pq-(p+q)]=o$
$\Rightarrow pq=p+q\Rightarrow \dfrac {p+q}{pq}=1$

$\Rightarrow \dfrac{1}{p}+\dfrac{1}{q}=1$
Multiple choice direction cosines and direction ratios three dimensional geometry maths

Determine if the points $(1,5)$ $(2,3)$ and $(-2,-11)$ are collinear.

  1. True

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

The given points are $A(1,5)$, $B(2,3)$ and $C(-2,-11)$.


Let us calculate the distance : $AB$, $BC$ and $CA$ by using distance formula.

$AB =\sqrt { (2-1)^{ 2 }+(3-5)^{ 2 } } =\sqrt { (1)^{ 2 }+(-2)^{ 2 } } $

$=\sqrt {1+4} = \sqrt{ 5 }$ units

$BC =\sqrt { (-2-2)^{ 2 }+(-11-3)^{ 2 } }=\sqrt { (-4)^{ 2 }+(-14)^{ 2 } }$

$=\sqrt {16+196} =\sqrt {212} = 2\sqrt{53}$ units

$CA =\sqrt { (-2-1)^{ 2 }+(-11-5)^{ 2 } }$

$=\sqrt { (-3)^{ 2 }+(-16)^{ 2 } } =\sqrt {9+256} = \sqrt {265 }$ 

$=\sqrt {5}\times\sqrt {53}$ units

From the above we see that : $AB+BC\neq CA$

Hence, the above stated points $A(1,5)$, $B(2,3)$ and $C(-2,-11)$ are not collinear.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

In each of the following find the value of $k$, for which the points are collinear.
(i) $(7,-2)$, $(5,1)$, $(3,k)$
(ii) $(8,1)$, $(k,-4)$, $(2,-5)$

  1. (i) $k = 4$
  2. (i) $k = 5$
  3. (ii) $k = 3$
  4. (ii) $k = 2$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

Since the given points are collinear, they do not form a triangle, which means area of the triangle is Zero.

Area of a triangle with vertices $({ x } _{ 1 },{ y } _{ 1 })$ ; $({ x } _{ 2 },{ y

} _{ 2 })$  and $({ x } _{ 3 },{ y } _{ 3 })$  is $ \left| \dfrac { {

x } _{ 1 }({ y } _{ 2 }-{ y } _{ 3 })+{ x } _{ 2 }({ y } _{ 3 }-{ y } _{ 1 })+{ x } _{

3 }({ y } _{ 1 }-{ y } _{ 2 }) }{ 2 }  \right| $


1) Substituting the points $({ x } _{ 1 },{ y } _{ 1 }) = (7,-2) $ ; $({ x

} _{ 2 },{ y } _{ 2 }) = (5,1) $  and $({ x } _{ 3 },{ y } _{ 3 }) = (3,k)$

In the area formula, we get

$ \left| \dfrac { 7(1-k) + 5(k+2) + 3(-2-1) }{ 2 }  \right|  =

0 $

$ \left| \dfrac { 7 -7k + 5k + 10 - 9 }{ 2 }  \right|  =

0 $

$ \left| \dfrac { 8 -2k }{ 2 }  \right|  =

0 $

$ \Rightarrow  8 - 2k = 0 $

$ \Rightarrow  k = 4 $

2) Substituting the points $({ x } _{ 1 },{ y } _{ 1 }) = (8,1) $ ; $({ x

} _{ 2 },{ y } _{ 2 }) = (k,-4) $  and $({ x } _{ 3 },{ y } _{ 3 }) = (2,-5)$ in the area formula, we get


$ \left| \dfrac { 8(-4+5) + k(-5-1) + 2(1+4) }{ 2 }  \right|  =

0 $

$ \left| \dfrac { 8 -6k +10 }{ 2 }  \right|  =

0 $

$ \left| \dfrac { 18 -6k }{ 2 }  \right|  =

0 $

$ \Rightarrow  18 - 6k = 0 $

$ \Rightarrow  k = 3 $

Multiple choice direction cosines and direction ratios three dimensional geometry maths

Are the points (1, 1), (2, 3) and (8, 11) collinear ?

  1. collinear

  2. Non collinear

  3. coplaner

  4. None of above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Area of triangle formed by these vertices is 
$\displaystyle \Delta =\frac { 1 }{ 2 } \begin{vmatrix} 1 & 1 & 1 \ 2 & 3 & 1 \ 8 & 11 & 1 \end{vmatrix}$
Applying ${ R } _{ 2 }\rightarrow { R } _{ 2 }-{ R } _{ 1 },{ R } _{ 3 }\rightarrow { R } _{ 3 }-{ R } _{ 1 }$
$\displaystyle \Delta =\frac { 1 }{ 2 } \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 0 \ 7 & 10 & 0 \end{vmatrix}=\frac { 1 }{ 2 } \left( 10-14 \right) =2$
Hence points are non collinear 

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

The straight lines
$\left.\begin{matrix}
2kx-2y+3=0\
x+ky+2=0\
2x+k=0
\end{matrix}\right}k\in R$  pass through the same point for

  1. no real value of $k$
  2. exactly one real value of $k$
  3. three real values of $k$
  4. all real values of $k$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given equation of lines passes through same point i.e. lines are concurrent
$kx-2y+3=0\ 
x+ky+2=0\ 
2x+k=0$
$\Rightarrow \left| \begin{matrix} 2k & -2 & 3 \ 1 & k & 2 \ 2 & 0 & k \end{matrix} \right| =0$
$\Rightarrow k^{3}-2k-4=0$
$\Rightarrow (k-2)(k^2+2k+2)=0$
The discriminant of the quadratic expression is negative.
Hence there is only one real value of $k$

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

The three distinct straight lines $ax+by+c=0$;$bx+cy+a=0$ and $cx+ay+b=0$ are concurrent then

  1. $a+b+c=0$
  2. $a^{3}+b^{3}+c^{3}=3 abc$
  3. $a=b=c$
  4. $a^{2}+b^{2}+c^{2}=ab+bc+ca$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

The lines are concurrent,
$=>\begin{bmatrix}a&b& c\ b &c &a\ c & a& b\end{bmatrix}=0$
Applying,$ C _{1}=C _{1}+C _{2}+C _{3}$,
$=\begin{bmatrix}a+b+c&b& c\ a+b+c &c &a\ a+b+c & a& b\end{bmatrix}$
Applying, $R _{1}=R _{1}-R _{2}$ and $R _{2}=R _{2}-R _{3}$,
$=(a+b+c)\begin{bmatrix}0&b-c& c-a\ 0 &c-a &a-b\ 1 & a& b\end{bmatrix}$
Expanding by $C _{1}$,
$=(a+b+c)(-a^{2}-b^{2}-c^{2}+ab+bc+ac)=0$
$=> (a+b+c)=0$
or,
$(a+b+c)(-a^{2}-b^{2}-c^{2}+ab+bc+ac)=0$
$abc-a^{3}-b^{3}+abc+abc-c^{3}=0$
$a^{3}+b^{3}+c^{3}=3abc$
So options are A and B.

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

If the points $(5, 5), (7, 7)$ and $(a, 8)$ are collinear then the value of a is

  1. $6$
  2. $3$
  3. $8$
  4. $7$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When three points are collinear, Slope of line joining any two points is same as the slope of line joining any other two points


Slope of line joining two points $ ({x} _{1}, {y} _{1}) $ and $ ({x} _{2}, {y} _{2}) $ is $\dfrac { {y} _{2} - {y} _{1}}{ {x} _{2} - {x} _{1}} $

So, Slope of line joining $ (5,5) ;  (7,7) $ is $ \dfrac {7-5}{7-5} = \dfrac {2}{2} = 1 $

And Slope of line joining $ (a,8) ;  (7,7) $ is $ \dfrac {7-a}{7-8} = a - 7 $

As they are collinear $ a - 7 = 1 => a = 8 $



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

Find the equation of a line whose inclination is $\displaystyle 30^{\circ}$ and making an intercept of -3/5 on the y-axis

  1. $ y = \dfrac {5}{\sqrt{6}}x +\dfrac {2}{5} $
  2. $ y = \dfrac {1}{\sqrt{3}}x -\dfrac {3}{5} $
  3. $ y = \dfrac {3}{\sqrt{7}}x -\dfrac {1}{3} $
  4. $\displaystyle \dfrac{-5}{3}x$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation of any straight line can be written as $ y = mx + c $, where $m$ is its slope and $c$ is its y - intercept.

As inclination is $ 30^o $, slope of the line $ =  tan (30 ^o) = \dfrac {1}{\sqrt{3}} $

So equation of line is $ y = \dfrac {1}{\sqrt{3}}x  -\dfrac {3}{5} $

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.