Question

What is the equation in standard form of a perpendicular line that passes through (5,-1) and what is the x-intercept of the line?

Answer 1

See below for steps to solve this kind of question:

Explanation:

Normally with a question like this we'd have a line to work with that also passes through the given point. Since we aren't given that, I'll make one up and then proceed on to the question.

Original Line (so called...)

To find a line that passes through a given point, we can use the point-slope form of a line, the general form of which is:

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

I'm going to set `m=2`. Our line then has an equation of:

`(y-(-1))=2(x-5)=>y+1=2(x-5)`

and I can express this line in point slope form:

`y=2x-11`

and standard form:

`2x-y=11`

For finding our parallel line, I'll use the point slope form:

`y=2x-11`

A perpendicular line will have a slope of `m_"perpendicular"=-1/m_"original"`

also known as the negative reciprocal.

In our case, we have the original slope as 2, so the perpendicular slope will be `-1/2`

With out slope and the point we want to go through, let's use the point slope form again:

`(y-(-1))=-1/2(x-5)=>y+1=-1/2(x-5)`

We can make this into standard form:

`y+1=-1/2x+5/2`

`1/2x+y=5/2-2/2`

`x+2y=3`

We can find the x intercept by setting `y=0`:

`x=3`

Graphically, it all looks like this:

original line:
graph{(2x-y-11)=0}

perpendicular line added:

graph{(2x-y-11)(x+2y-3)=0}