# 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?

Sep 29, 2017

$6 y - 8 x = 45$

#### Explanation:

Slope of the line segment $m 1 = \frac{y 2 - y 1}{x 2 - x 1}$
$m 1 = \frac{2 - 5}{1 - 5}$
$m 1 = - \frac{3}{-} 4 = \frac{3}{4}$
Mid point of the line segment $= \left(\frac{5 + 1}{2} , \frac{5 + 2}{2}\right)$
$= \left(3 , 3 \left(\frac{1}{2}\right)\right)$
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 $m 2 = - \left(\frac{1}{m} 1\right) = - \frac{1}{\frac{3}{4}} = - \frac{4}{3}$
Since the line segment is rotated by $\frac{\pi}{2}$ with mid point as the axis, the rotated line slope is $- \left(\frac{4}{3}\right)$ and the mid point coordinates $\left(3 , 3 \left(\frac{1}{2}\right)\right)$
Equation of the rotated line is
(y-y1)=m((x-x1)
$\left(y - 3 \left(\frac{1}{2}\right)\right) = - \left(\frac{4}{3}\right) \left(x - 3\right)$
$\frac{2 y - 7}{2} = \frac{- 4 x + 12}{3}$
$6 y - 21 = - 8 x + 24$
$6 y + 8 x = 45$