There is one other consideration for straight-line equations: finding parallel and perpendicular lines. Here is a common format for exercises on this topic:
They've given me a reference line, namely, 2x – 3y = 9; this is the line to whose slope I'll be making reference later in my work.
Content Continues Below
This line has some slope value (though not a value of "2", of course, because this line equation isn't solved for "y=").
I'll first need to find the slope of the reference line. I could use the method of twice plugging x-values into the reference line, finding the corresponding y-values, and then plugging the two points I'd found into the slope formula, but I'd rather just solve for "y=". (This is just my personal preference. If your preference differs, then use whatever method you like best.) So:
The first thing I'll do is solve "2x – 3y = 9" for "y=", so that I can find my reference slope:
So the reference slope from the reference line is .
Affiliate
Now I need to find two new slopes, and use them with the point they've given me; namely, with the point (4, –1). They want me to find the line through (4, –1) that is parallel to 2x – 3y = 9; that is, through the given point, they want me to find a line that has the same slope as the reference line. And they then want me to find the line through (4, –1) that is perpendicular to 2x – 3y = 9; that is, through the given point, they want me to find the line that has a slope which is the negative reciprocal of the slope of the reference line.
Since a parallel line has an identical slope, then the parallel line through (4, –1) will have slope . Hey, now I have a point and a slope! So I'll use the point-slope form to find the line:
This is the parallel line that they'd asked for, and it's in the slope-intercept form that they'd specified.
For the perpendicular line, I have to find the perpendicular slope. The reference slope is . For the perpendicular slope, I'll flip the reference slope and change the sign. Then the perpendicular slope is .
Again, I have a point and a slope, so I can use the point-slope form to find my equation. Note that the only change, in what follows, from the calculations that I just did above (for the parallel line) is that the slope is different, now being the slope of the perpendicular line.
Then the full solution to this exercise is:
parallel:
perpendicular:
Warning: If a question asks you whether two given lines are "parallel, perpendicular, or neither", you must answer that question by finding their slopes, not by drawing a picture! Pictures can only give you a rough idea of what is going on.
For instance, you would simply not be able to tell, just "by looking" at the picture, that drawn lines with slopes of, say, m_{1} = 1.00 and m_{2} = 0.99 are NOT parallel — and they'll sure as heck look parallel on the picture. But since 1.00 does not equal 0.99, the lines can not possibly be parallel.
In other words, to answer this sort of exercise, always find the numerical slopes; don't try to get away with just drawing some pretty pictures.
You can use the Mathway widget below to practice finding a parallel line through a given point. Try the entered exercise, or type in your own exercise. Then click the button to compare your answer to Mathway's. (The next widget is for finding perpendicular lines.) Or continue to the two complex examples which follow.
Please accept "preferences" cookies in order to enable this widget.
(Clicking on "Tap to view steps" on the widget's answer screen will take you to the Mathway site for a paid upgrade.)
Content Continues Below
You can use the Mathway widget below to practice finding a perpendicular line through a given point. Try the entered exercise, or type in your own exercise. Then click the button to compare your answer to Mathway's.
Please accept "preferences" cookies in order to enable this widget.
(Clicking on "Tap to view steps" on the widget's answer screen will take you to the Mathway site for a paid upgrade.)
Nearly all exercises for finding equations of parallel and perpendicular lines will be similar to, or exactly like, the one above. Here are two examples of more complicated types of exercises:
Since the slope is the value that's multiplied on "x" when the equation is solved for "y=", then the value of "a" is going to be the slope value for the perpendicular line. In other words, they're asking me for the perpendicular slope, but they've disguised their purpose a bit. It's up to me to notice the connection.
The first thing I need to do is find the slope of the reference line. I'll solve for "y=":
Then the reference slope is m = 9. The perpendicular slope (being the value of "a" for which they've asked me) will be the negative reciprocal of the reference slope.
I start by converting the "9" to fractional form by putting it over "1". Then I flip and change the sign. The result is:
Affiliate
The only way these two lines could have a distance between them is if they're parallel. (Otherwise, they must meet at some point, at which point the distance between the lines would obviously be zero.) Are these lines parallel? I'll solve each for "y=" to be sure:
...and:
The lines have the same slope, so they are indeed parallel. And they have different y-intercepts, so they're not the same line. Therefore, there is indeed some distance between these two lines. But how to I find that distance? It will be the perpendicular distance between the two lines, but how do I find that?
Affiliate
Advertisement
I know I can find the distance between two points; I plug the two points into the Distance Formula. But I don't have two points. Ah; but I can pick any point on one of the lines, and then find the perpendicular line through that point. Since the original lines are parallel, then this perpendicular line is perpendicular to the second of the original lines, too. Then I can find where the perpendicular line and the second line intersect. That intersection point will be the second point that I'll need for the Distance Formula.
I know the reference slope is . Then my perpendicular slope will be . Now I need a point through which to put my perpendicular line. I'll pick x = 1, and plug this into the first line's equation to find the corresponding y-value:
So my point (on the first line they gave me) is (1, 6). With this point and my perpendicular slope, I can find the equation of the perpendicular line that'll give me the distance between the two original lines:
Okay; now I have the equation of the perpendicular. The distance will be the length of the segment along this line that crosses each of the original lines.
Where does this line cross the second of the given lines? It'll cross where the two lines' equations are equal, so I'll set the non-y sides of the second original line's equaton and the perpendicular line's equation equal to each other, and solve:
The above more than finishes the line-equation portion of the exercise. I'll leave the rest of the exercise for you, if you're interested.
(To finish, you'd have to plug this last x-value into the equation of the perpendicular line to find the corresponding y-value. This would give you your second point. [It turns out to be , if you do the math.] Then you'd need to plug this point, along with the first one, (1, 6), into the Distance Formula to find the distance between the lines. [The distance turns out to be , or about 3.7442, if you plow through the computations.])
Note that the distance between the lines is not the same as the vertical or horizontal distance between the lines, so you can not use the x- or y-intercepts as a proxy for distance.
Of greater importance, notice that this exercise nowhere said anything about parallel or perpendicular lines, nor directed us to find any line's equation. It was left up to the student to figure out which tools might be handy. Don't be afraid of exercises like this. Yes, they can be long and messy. But even just trying them, rather than immediately throwing your hands up in defeat, will strengthen your skills — as well as winning you some major "brownie points" with your instructor.
URL: https://www.purplemath.com/modules/strtlneq3.htm
© 2018 Purplemath. All right reserved. Web Design by