Answer to Question #5949 in Algebra for Kurt
Write an equation in standard form for the line perpendicular to y=-1/2x+2 and passes through (6,-3).
Let's write the given equation in a form
x/2 + y = 2.
We see that the the direction vector of this line is
Let's seek the perpendicular line in a form
ax + by = c.
It's direction vector is (a,b). As these two lines are perpendicular, we have that
(1/2, 1)*(a,b) =& a/2 + b = 0 ==> b = -a/2
ax - ay/2 = c.
Let substitute point (6, -3):
6a& - a(-3)/2 = c
a:=1 ==> b = -1/2
6 + 3/2 = c
c = 7.5,
x& - y/2 = 15/2
and at last
2x - y = 15.
We have the equation of the line in standard form:
y = 2x - 15.