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