Line problemsDate: Thu, 1 Dec 1994 17:14:42 -0500 From: Anonymous Subject: Need Help in a hurry!!! Please help me to do these problems because I don't have the slightest idea how to do them. (1) Give the equation of a vertical line. (2) What is the equation of the vertical line containing the point 3,-4 (3) What is the eq. of the horizontal line containing the point 4,6 (4) What is the equation of the x-axis (5) What is the equation of the y-axis (6) Write the equations of 3 lines containing the point -1,1 (7) Does the point 2,4 lie on the line with equation x-5/y-3=3 (8) Give the ordered pairs of 3 points on the line with eq. y=3x/2+4 (9) Solve the equation 3x-5y=15 for y (10) State the x- and y-intercepts of the line with equation 5x-2y=10 (11) Solve for y, Y=4x+7y I would appreciate it if you would explain how to do these problems. Please write back soon. Date: Mon, 5 Dec 1994 15:57:50 -0500 (EST) From: Dr. Sydney Subject: Re: Need Help in a hurry!!! Hello there! Thanks for writing Dr. Math! I think I'll help you by going over some of the general concepts these questions are all about and working a few examples, and then maybe you can do the problems, okay? Okay, first let's talk about horizontal and vertical lines. A horizontal line will be of the form y = c while a vertical line will be of the form x = c for some constant, c. So, for instance, y = 28 is the horizontal line that is 28 units above the x axis. x = -6 is the vertical line that is 6 units to the left of the y-axis. Why do these equations describe horizontal/vertical lines? Well, let's break it down... say we take the equation y = 28. That equation is saying that given any point on the xy plane, if the y coordinate of that point is 28 than that point is on the graph of y=28. That means that x can be any number you want. So, for instance, (1, 28), (5, 28), (-898989898, 28) will all be on the graph of y = 28. If you draw a graph of this it will make more sense. Try plotting all points whose y coordinates are 28. You get a horizontal line, right? Can you apply the same ideas to vertical lines? Now, what do you know about the x-axis and y-axis? The x-axis is a horizontal line, right? So what might the equation for this line be? The y-axis is a vertical line. What might the equation for this line be? (What values do x and y have on the axes?) That should help with questions 4 and 5. Now, what if you are given an equation of a line and want to find points on that line? Your equation will give you a relation between 2 variables, x and y. So, if you let one of those variables be any number you want, and plug that number in for that variable you will have an equation with one variable. Solve that equation, and you have your point. For instance, say I want to know a point on the line 2x + 8y = 16. If I choose x = 0 (you can choose any number you want, but usually it is easy to choose 0), then the equation becomes 2(0) + 8y = 16. So we have: 8y = 16. Divide both sides by 8 to get: y = 2. So, when x = 0, y = 2. That means that the point (0,2) is on the line. Note we could have chosen any number for our x, but after we have chosen that number, there is only one possibility for what y can be. To check if a point is on a line, all you have to do is plug in values and check if everything looks all right. For instance, say I want to know if the point (1,2) is on the line with equation y = 3x - 1. Plug in x=1, y=2 and see what you get: So, does 2 = 3(1) - 1? Yes, so, the point (1,2) is on the graph. What about the point (0,1). Plug in x=0, y=1 and see what you get: Does 1 = 3(0) - 1? No, so the point (0,1) isn't on the graph. If you are given an equation with two variables and you are asked to solve for one of them, you have to play around with the equation until you have an answer. It is easiest to see how in an example. Say we want to solve the equation 2x + 8y = 16 for x. Then first you should get the x's on one side of the equals sign and the y's and constants on the other side of the equals sign. So, subtract 8y from both sides to get: 2x = 16 -8y. Now we are almost done...we want to know what x is (not what 2x is), so we divide both sides of the equation by 2 to get: x = (1/2)(16 - 8y) = 8 - 4y So, x = 8 - 4y We've solved the equation for x. So, the general strategy is to get all the x's on one side of the equation and then divide by a constant. On to your last question about intercepts. Do you know what an intercept is? An x-intercept is the point on the line where the line intersects the x-axis; a y-intercept is the point on the line where the line intersects the y - axis. We know that on the x-axis, y=0 and on the y-axis x=0, right? So, to find the x-intercept of an equation, set y=0 and solve for x. To find the y-intercept of an equation set x = 0 and solve for y. Remember the example I did above where I found that the point (0,2) was on the graph of 2x + 8y = 16? Well, (0,2) is a y-intercept for that graph. Another way to find intercepts is to first solve the equation for either variable, say you solve it for y. Then you have an equation of the form y = mx +b. Here, "m" is called the slope and "b" is the y-intercept.(If you plugged in x=0 to the equation, you would get y=b) I hope this helps with these problems. If you have any questions at all or want to check your work, feel free to write back. We were glad to hear from you! --Sydney, Dr. "Math Rocks!" |
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