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

Find the equations of the two straight lines drawn through the point $(0,a)$ on which the perpendicular let fall from the point $(2a,2a)$ are each of length $a$.
 then equation of the straight line joining the feet of these perpendiculars is $y+2x=5a$

  1. True

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

By calculating the lines passing through (0, a) and applying the condition that the perpendicular distance from (2a, 2a) is 'a', we can verify the equation of the line joining the feet of the perpendiculars.

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

The line $x+3y-2=0$ bisects the angle between a pair of straight lines of which one has equation $x-7y+5=0$. The equation of the other line is-

  1. $3x+3y-1=0$
  2. $x-3y+2=0$
  3. $5x+5y-3=0$
  4. $none$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
we have
$L _1 =x+3y-2=0---(1)$

$L _2=x-7y+5=0---(2)$

now,

we know that

family of line through the given lines is

$L=L _2+\lambda L _2=0$

$=x-7y+5+\lambda (x+3y-2)=0---(3)$

Distance of any point ray $(2,0)$ on the line $x+3y-2=0$ from the lime 
$x-7y+5=0$ and the line $L=0$ must be same
so,

$\Rightarrow \ \left |\dfrac {2+5}{\sqrt {50}}\right | = \left |\dfrac {2+2\lambda +5-2\lambda}{\sqrt {(1+\lambda)^2+(3\lambda -7)^2}}\right|$

$\Rightarrow \ \dfrac {7}{\sqrt {50}}=\dfrac {7}{\sqrt {(1+\lambda)^2 +(3\lambda -7)^2}}$

$\Rightarrow \ 10\lambda^2-40\lambda =0$

$10\lambda (\lambda -4)=0$

$\lambda =0,\ \lambda =4$

then, put $\lambda =4$ in equation $(3)$ and we get

$L=x-7y+5+4(x+3y-2)=0$

$L=x-7y+5+4x+12y-8=0$

$L=5x+5y-3=0$

Hence this is the answer.


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

If the straight line $2x+3y+1=0$ bisects the  angle between a pair of lines ,one of which in this pair is $3x+2y+4=0$, then the equation of the other line in that pair of line is 

  1. $3x+4y-9=0$
  2. $6x-7y-14=0$
  3. $9x+46y-28=0$
  4. $9x-23y-12=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The bisector of the angle between two lines is given by the angle bisector theorem. Given one line and the bisector, the other line can be determined by reflecting the known line across the bisector.

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

In the equation $2x^{2}+2hxy+6y^{2}-4x+5y-6=0$ represent a pair of straight lines then the length of intercept on the $x-$axis cut by the lines is

  1. $2$
  2. $4$
  3. $\sqrt {7}$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a pair of lines ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0, the intercept on the x-axis is found by setting y=0 and solving the resulting quadratic in x. The distance between the roots is the intercept length.

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) $\displaystyle 2fgh=ch^{2}+bg^{2}$