 There's a lot of information to unpack in this next example. Take a minute and read it. So the fact that we want to line in standard form means that the final equation will look something like this. Now in order to get to that point, in order to get to standard form, we're going to start with point-slope form. Now in order to use point-slopes form, we need a point and we need a slope. Well, we already know the point that this line will pass through. That's negative 1, 4. The line that's given to us has a slope of negative 6, and since we want something that's parallel to that line, parallel lines have the same slopes. So that means the slope that we'll use for point-slope will also be negative 6. So now we can work through the process of point-slope to slope intercept to standard form. So we'll start with point-slope form, substituting the values for the variables that we know. Now using some algebra, the distributive property will add for our job is to get this into slope intercept form. So first the distributive property, and then we'll add 4 to both sides. So now we have slope intercept form. Now all that's left to do is, again using some algebraic manipulation, get it into standard form. In order to do that, we'll add 6x to both sides. And so now we're finished. 6x plus y is equal to negative 2.