Algebra homework help?!?

Question

Find the equation in standard fform of the lines through point P that are (a) parallel to, and (b) perpendicular to, line L.

P(0,-2); L:y=x-3

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please explain to me how to do this

Answer 1

A) y=mx+b is the equation of a line
ax+by=c is the standard form
slope is 1 for the line and because the lines are paralell, they have the same slope. (since there is not number before x, you can put in a 1 and it will still be the same) Plug in the point to the equation.
-2=1*0+b
-2=0+b
-2=b (the y-interecept)
y=x-2 (subtract x from both sides)
-x+y=-2 (multiply the equation by -1)
x-y=2

B) Perpendicular lines have a negative reciprocal slope. So because the slop of the original line was 1, the slop of the line perpendicular to it is -1. Plug it in.
-2=-1*0+b
-2=b
y=-x-2 (add x to both sides)
x+y=-2

Answer 2

The slope of line L is 1 because that is the coefficient of x.

So a line || to L will also have a slope of 1 and, a line _|_ L will have a slope of -1

So equation of line || to L is y= x +b
Since this line goes through (0,-2) we have -2 =0x+b so b =-2
The equation is thus y= x-2 which is x-y-2=0 in standard form.

The equation of the _|_ line is y = -x+b so -2= b so y = -x-2, which in standard form is -x-y-2 =0 or x+y+2 = 0

Answer 3

a) To find the line that is parallel to L and passes through P, you first need to find your slope...

y=mx+b; m is slope.
L: y=(1)x-3
Your slope is 1.
Now the easiest thing to do next is to just do an equation.
Point-Slope Form is made for this...
(y-y1)=m(x-x1)... insert m (slope) and the points for P (0, -2)
(y+2)=1(x-0)
y+2=x
y=x-2
To change that into standard form...
y-x+2=0

b) Here you do the same thing, EXCEPT you must change your slope to its NEGATIVE INVERSE. To do that, you flip your slope... ( 1/1 changes to, you guessed it, 1/1.) Now make that negative... you get -1.
Let's go back to point-slope form.
(y-y1)=m(x-x1)
(y+2)=-1(x-0)
y+2=-x
y+x+2=0.