Help! Write the equation of the line that is perpendicular to the line y=2x+2 and passes through the point (6, 3).
Ok, we need to keep a couple things in mind with this one. The slope of a line that is perpendicular to another is it's negative inverse. What do you think will be the slope of the perpendicular line in this case?
I really don't understand slope, at all.
Slope is \(\displaystyle \frac{rise}{run}~or~\frac{\Delta y}{\Delta x}\) Make sense? A slope of \(\displaystyle \frac{5}{2}\) would mean that you would go up five units and right 2 units from your y-intercept.
Got it.
So, how would I find the slope in this case?
In this case we are given this equation, which is in slope-intercept form, \(y=2x+2\) It's basic form is \(y=mx+b\), where m is slope. What is the slope of this line?
Oh! 2!
Correct, now we need to find the slope of the line that is perpendicular. Which will be the negative reciprocal. For example, lets say we have a slope of \(\displaystyle \frac{a}{b}\). What would be the negative reciprocal of that?
negative a/b?
Or -a/b?
No, a reciprocal of a fraction is when you swap the numerator and the denominator. What do you think now?
b/a?
\(\displaystyle -\frac{b}{a}\) Now, we have a slope of \(\displaystyle 2~or~\frac{2}{1}\) What do you think that will be?
What do I think what will be?
The negative reciprocal of the slope from the given problem.
-1/2?
Correct! So the basic structure of our new line will be, \(\displaystyle y=-\frac{1}{2}+b\) Now to find b, the y-intercept, we need to plug in the points given. \((6,3)\rightarrow(x,y)\) Do you think you could do that? Solve for b with those givens?
Absolutely, however. Where do I put x? After -1/2?
\((\color{red}{6},\color{blue}{3})\rightarrow(\color{red}{x},\color{blue}{y})\) \(\displaystyle\color{blue}{y}=-\frac{1}{2}\color{red}{x}+b\) Make sense?
Yes. Thank you so much!
Very cool! Good job!
Ok, to answer your concerns, you PLUG IN THE VALUES into the equation with the +b on the end. And you solve for b and then replace it with it.
Oh, okay. Thanks!
Join our real-time social learning platform and learn together with your friends!