Question

Find the equation in standard form of lines P that are A) parallel to and B) perpendicular to line L P(1,2); L: 3x-2y=1P(8,7);L: y= -4

Answer 1

To find if two lines are parallel, the slope must be the same.

so m=m

for P(1,2); L: 3x-2y=1

First, solve the equation for y:

3x-2y=1

Subtract both sides by 3x

3x-2y=1

3x-3x-2y =1-3x

-2y=1-3x

Now, divide both sides by -2y

-2y/-2 = 1-3x

y =1/-2 +3x/2

The parallel line using the point P(1,2)

y-y1 =m(x-x1)

Replace the values and solve for y.

y-2=3x/2 -1

y=3x/2+2

So the parallel lines is y=3x/2+2

To find a perpendicular line, when you multiply the slopes the result must be equal to -1.

So:

m1*m2 = -1

Replace m1=3/2

m1*m2 = -1

3x/2* m2 = -1

m2 = -1/(3x/2)

m2 = -2/3

To find the line use:

y-y1 =m(x-x1)

y-2=-2/3(x-1)

y-2=-2x/3 +2/3

y= -2x/3 +8/3

So y= -2x/3 +8/3 is the perpendicular line.