Question 161715
First, manipulate x-2y=24 into the "slope-intercept" form of the line:
y = mx+b
where
m is slope
b is y-intercept
.
x-2y=24
substracting x from both sides:
-2y=-x+24
dividing both sides by -2:
y = (1/2)x - 12
Slope of the line is 1/2
.
For our line to be perpendicular, the product of our slope and 1/2 has to be -1:
Let m = slope of our new line
then
(1/2)m = -1
m = -2 
.
Recapping we now have the slope of our new line (-2) and a single point (1,3).  Plug this all into the "point-slope" form:
y - y1 = m(x-x1)
y - 3 = -2(x-1)
y - 3 = -2x+2
y = -2x + 5 (this is what they're looking for)