Find Equation of Line Perpendicular to x+y=0 Through (1,3)

[duki]

Hey guys, first week in Calc here. I need help with a problem.

It says to find the equations of the lines passing through (1,3) and having the following characteristics:

(the one I'm stuck on) perpendicular to the line x + y = 0

Could someone give me a hand? Maybe which formula to use?

 

[rocomath]

$$x+y=0$$

Solve for y. Lines perpendicular to the one you will solve for have different slopes and can be found with the equation $$m_1 m_2=-1$$

Point-slope equation ...

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

You should know these by heart.

 

[duki]

apologies for my ignorance, but I don't understand what you mean by the first part. Could you give me an example?

 

[rocomath]

Well I'm positive you can solve for y, so that's not what you're asking about.

Lines that are perpendicular to one another have different slopes. Their slopes can be found with the equation $$m_1 m_2=-1$$

$$y=-x$$

So $$m_1=-1$$

To find the slope of a line that is perpendicular to x+y=0, we just need to figure out what $$m_2$$ is.

 

[duki]

so $$y - 3 = -1 ( x - 1 )$$?

 

[rocomath]

$$m_1=-1$$ is the slope to your first line. They want the line perpendicular to that ... re-read what I wrote.

 

[duki]

oops, so $$y - 3 = 1 ( x - 1 )$$ ?

 

[rocomath]

oops, so $$y - 3 = 1 ( x - 1 )$$ ?

If that's the slope you found, yes.

You can check it by graphing it.