1. ## Equations of lines

I'm having trouble with some equations. I've tried to get the equation I need for these problems, but what I know how to do does not seem to be working.

1. Find the equation of the line that passes through (1/2,-2/3) and is perpendicular to the line 4x+8y=1. Give answer in point slope form.

Are there any tips on how to find perpendicular lines?

2. Find the slope and y-intercept of the line 6y+3=0.
the slope is zero
I'm confused as to how to find the y-intercept. I had been doing this, but it's not correct.
6y=-3
y=-3/6=-1/2

Thanks!!!

2. Originally Posted by lucy2284
I'm having trouble with some equations. I've tried to get the equation I need for these problems, but what I know how to do does not seem to be working.

1. Find the equation of the line that passes through (1/2,-2/3) and is perpendicular to the line 4x+8y=1. Give answer in point slope form.

Are there any tips on how to find perpendicular lines?
Hi Lucy,

First find the slope of the perpendicular by putting the equation in slope-intercept form ( $y=mx+b$).

$4x+8y=1$

$8y=-4x+1$

$y=-\frac{1}{2}x+\frac{1}{8}$

The slope is $-\frac{1}{2}$

Now the slope of a line perpendicular to this line will have a slope that is the negative reciprocal. So, the slope for our new line will be $2$.

Let substitute the point $\left(\frac{1}{2}, -\frac{2}{3}\right)$, and the slope, $2$, into the point slope form.

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

$y+\frac{2}{3}=2\left(x-\frac{1}{2}\right)$

Originally Posted by lucy2284
2. Find the slope and y-intercept of the line 6y+3=0.
the slope is zero
I'm confused as to how to find the y-intercept. I had been doing this, but it's not correct.
6y=-3
y=-3/6=-1/2

Thanks!!!
$y=-\frac{1}{2}$ graphs a horizontal line that crosses the y axis at $y=-\frac{1}{2}$, thus your y-intercept is $-\frac{1}{2}$. The slope of a horizontal line (line parallel to the x-axis) is always 0.