# Slope Intercept

1. Feb 7, 2010

### mistalopez

1. The problem statement, all variables and given/known data

Write the slope intercept forms of the equation of the lines through the given point (a) parallel to the given line and (b) perpendicular to the given line.

Point: (2,1) Line: 4x-2y=3

2. Relevant equations

y-y1=m(x-x1)

y=mx+b

3. The attempt at a solution

I first put the line into slope intercept form: y=-2x+(3/2)

a) Next, I used point-slope to get the parallel line:
y-y1=m(x-x1)
y-1=-2(x-2)
y-1=-2x+4
y=-2x+5
*However, the correct answer should be y=2x-3

b) I used the reciprical of the slope to make it perpendicular: m = 1/2
Next I used the point-slope to find the perpendicular line of the equation.
y-y1=m(x-x1)
y-1=(1/2)(x-2)
y-1=(1/2x)-1 *Multiply both sides by 2 to remove fraction
2(y-1)=((1/2x)-1)2
2y-2=1x-2
2y=1x
y=1x/2
*However, the correct answer in the book is y=(-1/2x)+2

What am I doing wrong?

2. Feb 7, 2010

### Mulder

it should be y = 2x - 3/2, you'll probably get the right answer with that

3. Feb 9, 2010

### gabrielh

Conveniently, one doesn't even have to use the point slope formula to solve this problem. By simply knowing that parallel lines have the same slope and perpendicular lines have a negative reciprocal slope and using the slope-intercept equation will yield you the proper result. For instance:

The slope intercept form of 4x - 2y = 3 would be:

y = 2x - $$3/2$$

For the parallel line, one would know that the slope must be the same as the given, thus, in this case, the slope would be 2. By using the given points, (2,1), one can determine the y-intercept of the parallel line like so:

1 = 2(2) + b

and solve for b.

The equation for the parallel line would then be y = 2x + whatever b would be.

The perpendicular line's equation is determined just the same. By knowing that the slope of a line perpendicular to the given is a negative reciprocal, one will see that the slope in this case would be $$-1/2$$ and thus:

1 = $$-1/2$$(2) + b

and solve for b once more.

I've always found this method quicker and easier for me than using the point-slope equation. Hopefully it will help you.