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
Answers
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)