2012-01-12T08:19:03-05:00
Write an equation in standard form for the line perpendicular to y=-1/2x+2 and passes through (6,-3).
1
2012-01-13T12:18:09-0500
Let's write the given equation in a form x/2 + y = 2. We see that the the direction vector of this line is (1/2, 1). 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 We obtain ax - ay/2 = c. Let substitute point (6, -3): 6a& - a(-3)/2 = c a:=1 ==> b = -1/2 Now, 6 + 3/2 = c and c = 7.5, so x& - y/2 = 15/2 and at last 2x - y = 15. We have the equation of the line in standard form: y = 2x - 15.
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