Question 182916
Start by putting:
6x-3y=5
into the "slope-intercept" form:
y = mx + b
where 
m is slope
b is y-intercept
.
6x-3y=5
-3y = -6x + 5
y = 2x - 5/3
So, now we know that the slope of this line is 2
.
If we want, a line that is perpendicular to the above, the slope has to be the "negative reciprocal"
So, our NEW line has to have a slope of -1/2 because:
2(-1/2) = -1
.
Recapping we have:
m = -1/2
and a single point at (4,10)
.
Stuff it all into the "point-slope" form:
y - y1 = m(x-x1)
y - 10 = (-1/2)(x-4)
y - 10 = (-1/2)x + 2
y = (-1/2)x + 12 (this is what they're looking for)