+816 Let $B$ be the reflection of $A$ over the line $y = \frac{1}{5} x + 3$. Find the coordinates of $B$ if $A = (1,4)$.
+129891 y= (1/5)x + 3
A= ( 1, 4)
The slope of the given line = (1/5)
Writing an equation for a perpendicular line with a slope of -5 passing through (1,4) we get
y = -5 ( x -1) + 4
y= -5x +9
Now, find the x coordinate of the intersection of these lines
(1/5)x + 3 = -5x + 9
(1/5)x + 3 = (-25/5)x + 9
(26/5)x = 6
x= 30/26 = 15/13
And y = (1/5)(15/13) + 3 = 3/13 + 3 = 42/13
So the intersection point is (15/13, 42/13)
From (1, 4) To (15/13, 42/13) we moved (15/13 - 1) = 2/13 on x and 42/13 - 4 = -10/13 on y
So...from the intersection point of th e lines we need to make the same moves
So B= (15/13 + 2/13 , 42/13 - 10/13) = ( 17/13, 32/13)
