Question

find the equation in point slope form of the lines through point P(2,0) that are a) parallel to, and b) perpendicular to the line of x-2y=3.

Answer 1

Parallel:
`y-0=(1)/(2)(x-2)`

Perpendicular:
`y-0=-2(x-2)`

Explanation:

Parallel lines have the same slope.

Perpendicular lines have opposite reciprocal slopes.

The point slope form of a line is:

`y-y_1=m(x-x_1)`

where
`(x_1,y_1)` is a point on the line and
`m` is the slope.

To find the slope we will first have to find the slope of the given line.

We are going to put it into slope-intercept form
`y=mx+b` because it tells us the slope,m, which is what we need.

Parallel lines will have the same slope,
`m`.

Perpendicular lines will have the opposite reciprocal slope:
`(-1)/(m)`.

So let's put
`x-2y=3` into
`y=mx+b` form by solving for
`y`.

`x-2y=3`

Subtract
`x` on both sides:

`-2y=-x+3`

Divide both sides by -2:

`y=(-x)/(-2)+(3)/(-2)`

Simplify:

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

or

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

So the slope is 1/2 for the given equation.

Our parallel line will also have slope 1/2.

Our perpendicular line will have slope -2.

Let's move onto finding these equations starting with the parallel one.

`y-y_1=m(x-x_1)`

Plug in
`m=(1)/(2)` and
`(2,0)`:

`y-0=(1)/(2)(x-2)`

Now the perpendicular one:

`y-y_1=m(x-x_1)`

Plug in
`m=-2` and
`(2,0)`:

`y-0=-2(x-2)`