A line segment with endpoints at (5, 5) and (1, 2) is rotated clockwise by pi/2. What are the new endpoints of the line segment?

1 Answer
sankarankalyanam
Sep 29, 2017

6y-8x=45

Explanation:

Slope of the line segment m1=(y2-y1)/(x2-x1)
m1=(2-5)/(1-5)
m1=-3/-4=3/4
Mid point of the line segment =((5+1)/2,(5+2)/2)
=(3,3(1/2))
When the line segment is rotated clock-wise by pi/2),
the rotated line becomes perpendicular to the existing line segment.
Slope of rotated line m2 = -(1/m1) = -1/(3/4)=-4/3
Since the line segment is rotated by pi/2 with mid point as the axis, the rotated line slope is -(4/3) and the mid point coordinates (3,3(1/2))
Equation of the rotated line is
(y-y1)=m((x-x1)
(y-3(1/2))=-(4/3)(x-3)
(2y-7)/2=(-4x+12)/3
6y-21=-8x+24
6y+8x=45

Related questions
See all questions in Rotations
Impact of this question
3131 views around the world
You can reuse this answer
Creative Commons License