# Thread: Find an equation for the line...

1. ## Find an equation for the line...

I am having trouble studying for my Algebra final. I am having problems giving equations for lines. The two problems that I am currently stuck on are--

(1) Give an equation for the horizontal through the point (2,3). (Is this wanting y=mx + b formula??)

and

(2) Find an equation for the line perpendicular to 2x + 5y = 10 and passes through (-2, 3) from (-1,3) (I think you do something with a negative reciprocal maybe, but I don't know where to begin)

Ashley

2. Originally Posted by aperkin2
I am having trouble studying for my Algebra final. I am having problems giving equations for lines. The two problems that I am currently stuck on are--
Originally Posted by aperkin2

(1) Give an equation for the horizontal through the point (2,3). (Is this wanting y=mx + b formula??)
Hi Ashley,

A horizontal line will always take the form y = b, where b is the y-intercept. Remember, the slope of a horizontal line is 0. Your y-intercept will be the y coordinate of your point.

Originally Posted by aperkin2

(2) Find an equation for the line perpendicular to 2x + 5y = 10 and passes through (-2, 3) from (-1,3) (I think you do something with a negative reciprocal maybe, but I don't know where to begin)
Here, you have to find the slope of the line defined by 2x + 5y = 10. A little manipulation of this equation will get it into the slope-intercept form:

$y=-\frac{2}{5}x+2$

Recognize that the slope is $-\frac{2}{5}$

The slope of a perpendicular will be the negative reciprocal of this slope. That is, we invert the fraction and change its sign. This slope becomes:

$m=\frac{5}{2}$

Now, this is where I'm having difficulty understanding which line you are trying to define.

What do you want to do? Are we defining a line perpendicular to 2x + 5y = 10 from (-2, 3) or are we defining a line perpendicular to 2x + 5y = 10 from (-1, 3)? Or is it something else entirely?