Home
Class 12
MATHS
Two particles start from point (2, -1), ...

Two particles start from point (2, -1), one moving two units along the line x+ y = 1 and the other 5 units along the line x - 2y = 4, If the particle move towards increasing y, then their new positions are:

Text Solution

AI Generated Solution

The correct Answer is:
To find the new positions of the two particles after they move along their respective lines, we will follow these steps: ### Step 1: Identify the initial position of the particles Both particles start from the point (2, -1). ### Step 2: Determine the direction and distance for the first particle The first particle moves 2 units along the line \(x + y = 1\). 1. **Find the slope of the line**: The equation \(x + y = 1\) can be rewritten as \(y = -x + 1\). The slope (m) is -1. 2. **Calculate the angle**: The angle \(\theta\) corresponding to this slope is given by: \[ \tan \theta = -1 \implies \theta = 135^\circ \text{ (in the second quadrant)} \] 3. **Calculate \(\cos \theta\) and \(\sin \theta\)**: \[ \cos \theta = -\frac{1}{\sqrt{2}}, \quad \sin \theta = \frac{1}{\sqrt{2}} \] 4. **Use parametric equations to find new coordinates**: The new coordinates \((x_1, y_1)\) after moving 2 units are given by: \[ x_1 = 2 + 2 \cdot \cos \theta = 2 + 2 \left(-\frac{1}{\sqrt{2}}\right) = 2 - \sqrt{2} \] \[ y_1 = -1 + 2 \cdot \sin \theta = -1 + 2 \left(\frac{1}{\sqrt{2}}\right) = -1 + \sqrt{2} \] ### Step 3: Determine the direction and distance for the second particle The second particle moves 5 units along the line \(x - 2y = 4\). 1. **Find the slope of the line**: The equation \(x - 2y = 4\) can be rewritten as \(y = \frac{1}{2}x - 2\). The slope (m) is \(\frac{1}{2}\). 2. **Calculate the angle**: The angle \(\theta\) corresponding to this slope is given by: \[ \tan \theta = \frac{1}{2} \implies \theta = \tan^{-1}\left(\frac{1}{2}\right) \] 3. **Calculate \(\cos \theta\) and \(\sin \theta\)**: Using the identity for \(\tan\): \[ \cos \theta = \frac{2}{\sqrt{5}}, \quad \sin \theta = \frac{1}{\sqrt{5}} \] 4. **Use parametric equations to find new coordinates**: The new coordinates \((x_2, y_2)\) after moving 5 units are given by: \[ x_2 = 2 + 5 \cdot \cos \theta = 2 + 5 \left(\frac{2}{\sqrt{5}}\right) = 2 + 2\sqrt{5} \] \[ y_2 = -1 + 5 \cdot \sin \theta = -1 + 5 \left(\frac{1}{\sqrt{5}}\right) = -1 + \sqrt{5} \] ### Final Results The new positions of the particles are: - First particle: \((2 - \sqrt{2}, -1 + \sqrt{2})\) - Second particle: \((2 + 2\sqrt{5}, -1 + \sqrt{5})\)

To find the new positions of the two particles after they move along their respective lines, we will follow these steps: ### Step 1: Identify the initial position of the particles Both particles start from the point (2, -1). ### Step 2: Determine the direction and distance for the first particle The first particle moves 2 units along the line \(x + y = 1\). ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • STRAIGHT LINES

    CENGAGE ENGLISH|Exercise CONCEPT APPLICATION EXERCISE 2.3|7 Videos
  • STRAIGHT LINES

    CENGAGE ENGLISH|Exercise CONCEPT APPLICATION EXERCISE 2.4|8 Videos
  • STRAIGHT LINES

    CENGAGE ENGLISH|Exercise CONCEPT APPLICATION EXERCISE 2.1|23 Videos
  • STRAIGHT LINE

    CENGAGE ENGLISH|Exercise Multiple Correct Answers Type|8 Videos
  • THEORY OF EQUATIONS

    CENGAGE ENGLISH|Exercise JEE ADVANCED (Numerical Value Type )|1 Videos

Similar Questions

Explore conceptually related problems

Find the distance of the line 4x+7y+5=0 from the point (1,""""2) along the line 2x-y=0 .

The number of points on the line x +y=4 which are unit distance apart from the line 2x+2y = 5 is :

The coordinates of the points on the circles x^(2) +y^(2) =9 which are at a distance of 5 units from the line 3x + 4y = 25 are :

If a ray travelling along the line x = 1 gets reflected from the line x + y = 1, then the eqaution the line along which the reflected ray travels is

Three particles start from the origin at the same time, one with a velocity v_(1) along the x-axis, second along the y-axis with a velocity v_(2) and third particle moves along the line x = y. The velocity of third particle, so that three may always lie on the same line is:

The triangle ABC, right angled at C, has median AD, BE and CF, AD lies along the line y = x + 3 , BE lies along the line y lies along the line y=2x +4 . If the length of the hypotenuse is 60, fin the area of the triangle ABC (in sq units).

A particle from the point P(sqrt3,1) moves on the circle x^2 +y^2=4 and after covering a quarter of the circle leaves it tangentially. The equation of a line along with the point moves after leaving the circle is

A particle moves along the curve y=x^2 + 2x . At what point(s) on the curve are x and y coordinates of the particle changing at the same rate?

A particle of mass m is moving along the line y-b with constant acceleration a. The areal velocity of the position vector of the particle at time t is (u=0)

Particle moves from point A to point B along the line shown in figure under the action of force. VecF = - x hati + y hatj . Determine the work done on the particle by vec F in moving the particle from point A to point B.