Tag: different forms of equation of a line

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Questions Related to different forms of equation of a line

Multiple choice maths functions and graphs different forms of equation of a line

The direction with $+x-axis$ in which a straight line will be drawn through the point $\left(1,2\right)$ so that its point of intersection with the line $x+y=4$ may be at a distance $\sqrt { \dfrac { 2 }{ 3 }  }$ from the point $\left(1,2\right)$ can be:

  1. ${15}^{o}$
  2. ${30}^{o}$
  3. ${45}^{o}$
  4. ${75}^{o}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving the geometry: The line through (1,2) at angle θ to +x-axis meets x+y=4 at distance √(2/3). Using parametric form and distance formula gives θ = 15° as the valid solution. This involves setting up the line equation, finding intersection point, and applying distance constraint.

Multiple choice maths functions and graphs different forms of equation of a line

The curve satisfying the equation $\dfrac { dy }{ dx } =\dfrac { y(x+{ y }^{ 3 }) }{ x({ y }^{ 3 }-x) } $ and passing through the point (4, -2) is

  1. ${ y }^{ 2 }=-2x$
  2. ${ y }=-2x$
  3. ${ y }^{ 3 }=-2x$

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

This is a homogeneous differential equation. Rewrite as dy/dx = (yx + y⁴)/(xy³ - x²). Using the substitution y = vx, we get v + x(dv/dx) = v(x + vx³)/(x(v³x³) - x²) = v(1 + v³x²)/(v³x² - 1). Solving this leads to the family of curves y³ = kx. Using point (4,-2): (-2)³ = k(4), so k = -8/4 = -2. Therefore y³ = -2x.

Multiple choice maths functions and graphs different forms of equation of a line

if the equation ${ 4x }^{ 2 }+2\sqrt { 3xy } +{ 2y }^{ 2 }-1=0$ becomes ${ 5x }^{ 2 }+{ y }^{ 2 }=1,\quad$  when the axes are rotar trough an angle ${ 45 }^{ 0 }$ , then the original  equation of the curve  is :'

  1. ${ 15 }^{ 0 }$
  2. ${ 30 }^{ 0 }$
  3. ${ 45 }^{ 0 }$
  4. ${ 60 }^{ 0 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths functions and graphs different forms of equation of a line

The equation of a straight line passing through a point $(-5,4)$ and which cuts off an intercept of $\sqrt{2}$ units between the lines $x+y+1=0$ and $x+y-1=0$ is

  1. $x-2y-13=0$
  2. $2x-y+14=0$
  3. $x-y+9=0$
  4. $x-y+10=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the given point  be $A=(−5,4)$ and the given lines be $l _1 \rightarrow x+y+1=0$ and  $l _2\rightarrow x+y-1=0$

The  point $A$ lies on $l _1$


If segment $AM\perp l _2$ and $M$ lies on $l _2$, then, the distance $ AM$ is given by,


$\Rightarrow AM=\dfrac{|−5+4−1|}{\sqrt{1^2+1^2}}=\dfrac{2}{\sqrt 2}=\sqrt 2$


$\Rightarrow $ This means that if $B$ is any point  on $l _2$ then  $AB>AM$. No line other than $AM$ cuts off an intercept of

length $\sqrt 2$ between $l _1$ and $l _2$.


$\Rightarrow $To determine the equation of $AM$, we need to find the co-ordinates of the Point $M$


Since, $AM\perp l _2$ and  the slope $l _2$ is $−1$, the slope of$AM$ must be $1$.  Also  $A(−5,4)$ lies on $AM$


By the point slope formula, the equation of the required line is


$\Rightarrow  y−4=1(x−(−5))$

$\Rightarrow y-4=x+5$

$\Rightarrow x−y+9=0$

Multiple choice maths functions and graphs different forms of equation of a line

Equation of a straight line passing through the point $(4, 5)$ and equally inclined to the lines $3x=4y+7$ and $5y=12x+6$ is?

  1. $9x-7y=1$
  2. $9x+7y=71$
  3. $7x+9y=73$
  4. $7x-9y+17=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The slopes of the given lines are m₁ = 3/4 and m₂ = 12/5. For a line to be equally inclined to both, its slope m must satisfy |(m-m₁)/(1+mm₁)| = |(m-m₂)/(1+mm₂)|. Solving gives two possible slopes: m = -7/9 (internal bisector) or m = 9/7 (external bisector). Using point (4,5) with slope -7/9: y-5 = (-7/9)(x-4), giving 9y-45 = -7x+28, or 7x+9y=73.

Multiple choice maths functions and graphs different forms of equation of a line

The number of values of $c$ such that the straight line $y=4x+c$ touches the curve $x^{2}+4y^{2}=4$, is

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

Substituting y = 4x + c into x² + 4y² = 4 gives: x² + 4(4x+c)² = 4, or x² + 4(16x² + 8cx + c²) = 4. This simplifies to 65x² + 32cx + 4c² - 4 = 0. For the line to be tangent to the ellipse, this quadratic must have exactly one solution, so discriminant = 0: (32c)² - 4(65)(4c²-4) = 0. This gives 1024c² - 1040c² + 1040 = 0, or -16c² + 1040 = 0, giving c² = 65. Therefore c = ±√65, so there are 2 values.

Multiple choice maths functions and graphs different forms of equation of a line

The equation of a straight line passing through the point (-5,4) and which cuts off an intercept of $\sqrt { 2 } $ unit between the lines $x+y+1=0$ and $x+y-1=0$ is:

  1. $2x-y+14=0$
  2. $3x+y+11=0$
  3. $x-y+9=0$
  4. $4x+5y=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the given point be $A=(-5,4)$ and the given lines be,

$L _1:x+y+1=0$ and 
$L _2:x+y-1=0$
Observe that, $A\in L _1$.
If segment $AM\perp L _2,$ $M\in L _2,$ then, the distance $AM$ is given  by,

$\Rightarrow$  $AM=\dfrac{|-5+4-1|}{\sqrt{1^2+1^2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}$

This means that if $B$ is any point on $L _2,$ then, $AB>AM.$
In other words, no line other than $AM$ cuts off an intercept of length $\sqrt{2}$ between $L _1,$ and $L _2$ or $AM$ is the required line.
To determine the equation of $AM,$ we need to find the co-ordinates of the point $M$.
Since, $AM\perp L _2,$ and the slope of $L _2$ is $-1,$ the slope of $AM$ must be $1.$
Further, $A(-5,4)\in AM$
By the slope-point form the equation of the required line is,
$\Rightarrow$  $y-4=1(x-(-5))$
$\Rightarrow$  $y-4=x+5$
$\Rightarrow$  $x-y+9=0$

Multiple choice maths functions and graphs different forms of equation of a line

The equation of a straight line passing through the point (-5,  4) and which cuts off in intercept of $\sqrt { 2 } $ unit. between the lines $x+y+1=0$ and $x+y-1=0$ is:

  1. $2x-y+14=0$
  2. $3x+y+11=0$
  3. $x-y+9=0$
  4. $4x+5y=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the given point be $A\left( {{x} _{1}},{{y} _{1}} \right)=\left( -5,4 \right)$ and the given lines be

$ x+y+1=0\,\,......\,\,\left( 1 \right) $

$ x+y-1=0\,\,......\,\,\left( 2 \right) $

Observe that,

From equation (1) to,

Let AM is perpendicular distance

Then, $AM=\dfrac{\left| -5+4-1 \right|}{\sqrt{{{1}^{2}}+{{1}^{2}}}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}$

Let B is any point on equation $(2)$to,

From equation (2)

$ x+y-1=0 $

$ y=-x+1 $

On comparing that,

$y=mx+c$

Then, $m=-1$

Slope of perpendicular line is

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

Then, equation of line is

$ y-{{y} _{1}}=m\left( x-{{x} _{1}} \right) $

$ \Rightarrow y-4=1\left( x+5 \right) $

$ \Rightarrow y-4=x+5 $

$ \Rightarrow x-y+5+4=0 $

$ \Rightarrow x-y+9=0 $

Hence, this is the answer.

Multiple choice maths functions and graphs different forms of equation of a line

A line passing through the points of intersection of $x+y=4$ and $x-y=2$ makes an angle $\tan^{-1}(3/4)$ with the x-axis. It intersects the parabola $y^2=4(x-3)$ at points $(x _1, y _1)$ and $(x _2, y _2)$ respectively. Then $|x _1-x _2|$ is equal to?

  1. $\dfrac{16}{9}$
  2. $\dfrac{32}{9}$
  3. $\dfrac{40}{9}$
  4. $\dfrac{80}{9}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving x+y=4 and x-y=2 gives the intersection point (3,1). A line through this point with slope 3/4 has equation: y-1 = (3/4)(x-3), or 4y-4 = 3x-9, or 3x-4y=5. Substituting into y²=4(x-3) gives a quadratic. The x-coordinates of the intersection points are found by solving for the parameter along the line. The difference |x₁ - x₂| = 16/9.