# Thread: more algebra 1 help

1. ## more algebra 1 help

I need help with this problem: "Find the equation of a line that is perpendicular to y = 6-3x and passes through the x-axis at -4." I keep getting y= 1x + 4 , which I know is wrong because the slope is supposed to be 1/3.

2. ## Re: more algebra 1 help

Originally Posted by ineedhelpwithmyhomework
I need help with this problem: "Find the equation of a line that is perpendicular to y = 6-3x and passes through the x-axis at -4." I keep getting y= 1x + 4 , which I know is wrong because the slope is supposed to be 1/3.
You are correct that the slope must equal $\dfrac{1}{3}.$

So $y = m + \dfrac{1}{3} * x.$

What does it mean that the line passes through the x-axis at - 4? It means that y = 0 when x = - 4.

$0 = m + \dfrac{1}{3} * (-\ 4) \implies m = -\ \dfrac{1}{3} * (-\ 4) = \dfrac{4}{3}.$

$\therefore y = \dfrac{4}{3} + \dfrac{1}{3} * x.$

Does that equation have the right slope? Does it cross the x-axis at x = - 4?