# Find an equation for the line with y-intercept 3 that is perpendicular to the line y= (2/3)x-4

*print*Print*list*Cite

We know that the y-intercept is 3

Then we have the point (0,3) passes through the line

Now we also know that the line is perpendicular to y=(2/3)x-4

Then the slope is (-3/2) (flip and change sign)

The the equation is:

y-y1= m(x-x1)

y-3= (-3/2)(x-0)

y= (-3/2)x +3

First, let's recall the fact that 2 line are perpendicular if the product of their slopes is -1.

Because the given equation y = (2/3)x-4 is written in the standard form, namely y = mx+n, the slope is m1 = 2/3.

The product of the slopes of the perpendicular lines is:

m1*m2 = -1

(2/3)*m2 = -1

We'll divide by (2/3):

m2 = -1/(2/3)

**m2 = -3/2**

The coordinates of the intrsection point with y axis are:(0,3).

The equation of the line that passes through a given point and it has a known slope is:

y - 3 = (-3/2)(x-0)

y - 3 = -3x/2

We'll add 3 both sides:

**y = (-3/2)*x + 3**

Any line which makes an intercept of 3 on y axis is of the form y = mx+3, where m is the slope of the lthe line

y =(2/3)x-4 is aline which has the slope (2/3)

Now y = mx+3 is perpendicular to the line y = (1/2)x -4 , if the product of the slopes of these two lines is -1.

So m*(2/3) = -1. This determines m = -1*(3/2) -3/2

So the slope of the line y= mx+3 is (-3/2). Or y=(3/2)x+3 is the line perpendicular to y = (2/3)x - 4