Find the equation of the perpendicular line of 3y - 6x + 9 = 0 and passes through the origin.
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
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
y = -0.5x