 This last example asks us for a line in slope-intercept form that passes through a given point, 2, negative 3, and is perpendicular to this given line, y equals 1 third x plus 1. So in order to solve this problem, we're still going to use this previous kind of algorithm, these steps, start with point-slope form, go to slope-intercept. However, since we don't need standard form, we're just going to stop when we get to slope-intercept form. So let's begin with point-slope form. So of course the given point is 2, negative 3. The slope that's given to us is in this line, we have 1 third is the given slope. However, we want a line that's perpendicular to that slope. And perpendicular slopes are opposite reciprocals. And so what that means is, since 1 third is positive, the perpendicular slope will be negative. And 1 third, flip that over, becomes positive 3. So our perpendicular slope is negative 3. So then we'll use that slope and the point 2, negative 3, and we'll stick that into point-slope form. Pardon me. So point-slope form. So point-slope says y minus y1, and y1 is negative 3, so minus negative 3 is like plus 3, times the slope of negative 3 times the quantity x minus 2. And now we need to change that into slope-intercept form. We'll use the distributive property first, negative 3 times x, and negative 3 times negative 2. And that yields y plus 3 is equal to negative 3x plus 6. And then lastly, we'll subtract 3 from both sides of the equation. And there we go. All done. Start with point-slope, and then change it into slope-intercept form.