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

Answer:

Parallel: $y-0=\frac{1}{2}(x-2)$

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

Step-by-step 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: $\frac{-1}{m}$.

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:

$y=\frac{-x}{-2}+\frac{3}{-2}$

Simplify:

$y=\frac{x}{2}-\frac{3}{2}$

or

$y=\frac{1}{2}x-\frac{3}{2}$.

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 $m=\frac{1}{2}$ and $(2,0)$:

$y-0=\frac{1}{2}(x-2)$

Now the perpendicular one:

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

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

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