Slopes, and more slopes.... :-( I need help.?

Question

A line passing through (-6, 4) and (3, y) is perpendicular to a line with slope 3/2. Find the value of y.

Answer 1

Perpendicular lines have negative reciprocal slopes.

Since line you are looking for is perpendicular to a line with a slope of 3/2, this means that the line you are searching for has a slope of  

(-2/3).

Slope-Intercept form:  y=mx +b

Slope= change in y / change in x = (y2-y1)/ (x2-x1) (the numbers 1 and 2 are subscripts)

We know that the slope is -2/3.

So:

(-2/3)= (y - 4) / (3-(-6))

(-2/3)=(y - 4) / (3+6)

(-2/3) = (y - 4) / (9)

Multiply both sides by 9

-6=y - 4

Add 4 to both sides to solve for y.

-2=y

Answer:  The value of y is (-2).

Answer 2

the slope of (-6, 4 ) and (3 , y )

is　4-y / -6-3

= 4-y/-9

= 3/2

4-y / - 9 = 3 / 2

- 2 ( 4-y )  = 27

-8+2y = 27

2y = 35

y = 35/2

　

Answer 3

previous answer has gone wrong.

To be perpendicular to 3/2,   the slope must be -2/3

Now solve for y as before.

Answer 4

Perpendicular lines have the slope of negative reciprocal:

m = -(x2-x1)/(y2-y1) for perpendicular lines

m = 3/2 = -(3- (-6))/(y-4)

3/2 = -(9/(y-4))

3/2 = -9/(y-4)

(3/2)(y-4) = -9

(3/2)y - 6 = -9

(3/2)y = -3

y = -3(2/3)

y = -2