Writing equations of lines parallel and perpendicular to a given line through a point

Question

Consider the line 2x + 6y = -4

Find the equation of the line that is perpendicular to this line and passes through the point (−5,  3)

Find the equation of the line that is parallel to this line and passes through the point (−5,  3)

Answer 1

To answer both questions I am going to utilize point slope form. Given any point (x₁,y₁) and slope m, we can write the equation of a line as:

y - y₁ = m(x - x₁)

We already were given a point (-5, 3) so now all we need is an appropriate slope for each part. We need to know that parallel lines have the same slope, and that perpendicular lines have opposite, reciprocal slopes (for example, 3/4 and -4/3 would be perpendicular slopes).

Now we just have to find the slope of the line that was given to us: 2x + 6y = -4

We can do this by rearranging the equation and putting it in slope-intercept form(y = mx + b) where m is the slope.

We can subtract 2x from both sides to get:

6y = -2x - 4

Then divide both sides by 6 to get:

y = -2/6 x - 4/6

Reducing the fractions

y = -1/3 x - 2/3 The important piece here is that the slope of this line is -1/3

So to write our parallel line we will use the slope of -1/3 and the point (-5, 3) and to write our perpendicular line we will use a slope of 3/1(or just 3) and the point (-5, 3)

So the perpendicular line could be written:

y - -5 = 3(x - 3) which simplifies to y + 5 = 3(x - 3)

Our parallel line would look like:

y - -5 = -1/3(x - 3) which simplifies to y + 5 = -1/3(x - 3)

There you go! If you have to go further and write these equations in y = mx + b form you can just distribute and solve for y.