Question

8.
At some instant, a particle is moving along a straight line 2x-3y = 2 and its co-ordinates
on that line are (4, 2). Now at another instant the same particle is moving along a
straight line 3x+4y = 7 and its co-ordinate are (1, 1). The co-ordinates of the axis about
which it is in pure rotation are​

Answer 1

Answer:

Since the particle is moving in a straight line at the given instants it must be moving tangent to the circle of rotation,

y1 = 2/3 x1 - 2/3    standard form of first equation where y = m x + b

A line perpendicular to this line (a radius) will have the form

Y1 = M X1 + B    where m * M = -1       condition for perpendicular lines

So 2 = -3/2 * 4 + B     using the coordinates of the intersection

B = 8    and the equation of the line along the radius is

Y1 = -3/2 X1 + 8

Using the same procedure for the second line

Y2 = 4/3 X2 - 1/3      since y2 = -3/4 x2 + 7/4 passing thru (1, 1)

The radii will intersect at the center of rotation so Y1 = Y2 and X1 = X2

-3/2 X + 8 = 4/3 X - 1/3

Solving for X:         X = 50/17

Substituting X in either equation for Y gives Y = 61/17

So the coordinates of the center of rotation are (50/17, 61/17)