MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
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MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
Since lines that are perpendicular to each other have slopes that are negative reciprocals of each other, you need the slope of one line to find the slope of the line perpendicular to that line. When given an equation of a line in standard form, rewrite it using inverse operations so that the line is in slope-intercept form so you can easily see what is the slope. Once you know the slope of the first line, find the negative inverse of the slope to find the slope of the line perpendicular to the first line. Then use point-slope form and plug in the slope and coordinates of the given point to find the equation of the line perpendicular to the original line. Solve this equation for y so that it is in slope-intercept form.
This problem here asks me to find the equation of a line perpendicular to a given line through a given point, Sounds pretty straight forward, except for this one small problem. This equation is not in y equals mx plus b form. That means it’s going to be hard for me to find the slope. Before I can do this problem and what I want to do is use this equation, I have to find the slope.
So first thing I’m going to do is work with this guy, manipulate it around until I can get it into y equals mx plus b form to find the slope. So I want to get y all by itself. I’m going to go through, let me use a different colour and subtract 3x from both sides so that I have 2y equals -3x plus 4. I still need to get y all by itself so divide by 2, and I’ll have y equals -3/2x plus 2.
Now I’m in good shape because I can find the slope and by the way I’m done with this equation now. I just kind of used it to find this slope. I’m done with it. I don’t care anything else about this equation. I don’t care about the y intercept, I’m not going to have to graph it. I can forget about it. All I needed to find was that slope -3/2.
Okay, so for my new line, I don’t want to use this exact same slope because the lines are perpendicular. Perpendicular means opposite sign, reciprocal slopes, so my new slope is going to be +2/3, opposite sign, reciprocal. There it is right there, it goes in there for m. Everything else is pretty straight forward, y minus y1 equals my new slope number times x take away x1, watch out for that minus, minus when you’re simplifying. 2/3 that becomes x plus 1, go ahead and distribute that 2/3, 2/3x plus 2/3 and then the next thing I’m going to have to do, after I bring it over here is find the common denominator for when I’m adding 2 to both sides. Adding 2, adding 2, so when I’m adding 2 to 2/3, I need to think about a common denominator. Instead of 2, I could write this as 6/3, so when I’m adding, I’m going to be doing 2/3 plus 6/3, to give me 8/3 as my intercept.
That’s a pretty nasty problem not only because you had to find the slope first before you could do any of this business, but also because you had to deal with some fractions and common denominators. So if you guys can do this problem, and you understand everything, you’re at the A+ level, you’re ready to move on. If you’re not quite there yet, that’s okay, don’t be too hard on yourself, hopefully you guys can understand the general processes so that you can do these problems on your own.
Unit
Linear Equations and Their Graphs