# Math Help - Find the equation of a line that is perpendicular to another.

1. ## Find the equation of a line that is perpendicular to another.

Okay, so I am given that the the equation passes through the point (-4, 2) and is perpendicular to 10x - 2y - 6 = 0. Dolefully I have done the calculations, over and over, producing the same results. I put the equation into slope-intercept and procure y = 5x -3
So, m_1 = 5 and m_2 = -1/5. I then take the new slope and use the point-slope equation and, after all of the calculations, find myself with the resultant: 5x + y + 18 = 0
(they ask it to be formatted into general form). The true answer, according the answer key, is x + 5y - 6 = 0--this can not be true!

2. ## Re: Find the equation of a line that is perpendicular to another.

The slope of the line which you're searching is indeed $\frac{-1}{5}$ that means we have untill now the equation: $y=\frac{-1}{5}x+q$
We have to find q this can be done by substituting the given point into the equation:
$2=\frac{4}{5}+q \Leftrightarrow q=2-\frac{4}{5} \Leftrightarrow q=\frac{6}{5}$
Therefore the equation of the straight line becomes:
$y=\frac{-1}{5}x+\frac{6}{5}$
$\Leftrightarrow 5y=-x+6$
$\Leftrightarrow 5y+x-6=0$

3. ## Re: Find the equation of a line that is perpendicular to another.

Why wouldn't the point-slope formula work?

4. ## Re: Find the equation of a line that is perpendicular to another.

y = -x/5 + b ; substitute -4,2:
b = 6/5

y = -x/5 + 6/5
Take over...

5. ## Re: Find the equation of a line that is perpendicular to another.

If you mean:
$y-2=\frac{-1}{5}\left(x+4)$
$\Leftrightarrow y-2=\frac{-x}{5}-\frac{4}{5}$
$\Leftrightarrow 5y-10=-x-4$
$\Leftrightarrow 5y+x-10+4=0$
$\Leftrightarrow 5y+x-6=0$

6. ## Re: Find the equation of a line that is perpendicular to another.

Well, what formula is this then:

m= (x - x_1)/(y - Y_1)?

7. ## Re: Find the equation of a line that is perpendicular to another.

Originally Posted by Bashyboy
Why wouldn't the point-slope formula work?
It will...go study:
Straight-Line Equations: Point-Slope Form