# Find the equation of the perpendicular line of 3y - 6x + 9 = 0 and passes through the origin.

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Let the equation of the line be :

y- y1 = m ( x-x1) where ( x1, y1) any point passes through the line and m is the slope.

Given that the line passes through the origin point

Then the line passes through the point ( 0, 0)

Let us substitute:

==> y- 0 = m (x- 0)

==> y= mx

Now we will find the slope:

Given the perpendicular line :

3y - 6x + 9 =0

We will write in slope form:

==> 3y = 6x - 9

==> y= 2x - 3

Then the slope = 2

Then the perpendicular slope = -1/2

Then :

** y= (-1/2) x**

Two lines are perpendicular when the product of their slopes is -1. For this reason, we'll write the given equation in the point slope form.

y = mx + n

3y - 6x + 9 = 0

We'll isolate 3y to the left side:

3y = 6x - 9

We'll divide by 3:

y = 2x - 3

Comparing the equations put in the point slope form, we'll have the slope of the given line: m1 = 2.

We also know that:

m1*m2 = -1, where m2 is the slope of the perpendicular line.

m2 = -1/m1

m2 = -1/2

The equation of the perpendicular line, that passes through the origin and has the slope m2 is:

y - 0 = m2(x - 0)

y = -x/2

or

**y = -0.5x**