# Answer to Question #5949 in Algebra for Kurt

Question #5949

Write an equation in standard form for the line perpendicular to y=-1/2x+2 and passes through (6,-3).

Expert's answer

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.

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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