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# Writing equations of perpendicular lines (example 2)

CCSS Math: HSG.GPE.B.5

## Video transcript

- [Instructor] Find the equation of a line perpendicular to this line that passes to the point two comma eight. So this first piece of information that it's perpendicular to that line right over there. What does that tell us? Well if it's perpendicular to this line, it's slope has to be the negative inverse of two-fifths. So its slope, the negative inverse of two-fifths, the inverse of two-fifths is five. Let me do it in a better color. A nicer green. If this lines slope is negative two-fifths, the equation of the line we have to figure out that's perpendicular, the slope is going to be the inverse. So instead of two-fifths, it's gonna be five halves. And instead of being a negative, it's going to be a positive. So this is the negative inverse of negative two-fifths, right. You take the negative sign, it becomes positive. You swap the five and the two, you get five halves. So that is going to have to be our slope. And we can actually use the point slope form right here. It goes through this point right there. So let's use point slope form. Y minus this Y value which has to be on the line. Is equal to our slope, five halves times X minus this X value. The X value when Y is equal to eight. And this is the equation of the line in point slope form if you wanna put it in slope intercept form. You can just do a little bit of algebra. Algebraic manipulation. Y minus eight is equal to let's distribute the five halves. So five halves X minus five halves times two is just five. And then add eight to both sides. You get Y is equal to five halves X. Add eight to negative five. So plus three. And we are done.