
Algebra
I have three questions.
1. Write the equation of a line that passes through the indicated points: O(2,3) and P(1,1)
2. Find the equation of a line that passes through (2,3) and parallel to the line whose equation is 2x  5y = 6.
3. Given the points P(1,1), Q(4,2), and R(6,4), a. Find the slopes of line segments PQ and PR. b. From your answer to part a, what can you conclude regarding the points P, Q, and R? c. Find the equation of the line that passes through P and R. d. Does the point S(10,8) lie on the line PR? Why or why not?
(The underlined letters are supposed to be lined on top.)
Re: Algebra
For #1, notice that the slope or m number in y = mx + b can be calculated by going from one point to the other, and it doesn't matter which one you choose I guess. So the fraction 31 over 2  1 is the same as 1  3 over 1  2. In case you didn't follow that, you take the last numbers in the coordinates and subtract them and make that the top of the fraction for the m number, and you take the first numbers of the coordinates and subtract them for the bottom of the fraction. This fraction is the m number, which is the slope of the line, or the tangent of the angle with the horizontal.
igloo myrtilles fourmis
Re: Algebra
problem #1 continued...
Now you know the slant of the line is up and to the right because m positive. It slants sort of at 2 o'clock. This is a rough guess, but the actual number for m is twothirds. m = (2/3) Now you can plug this number m and an x,y point into the y = mx + b equation and solve for b. Then when you know b's value, then the equation will be y=#x + #, where # is two different numbers. The y and the x stay in the final answer.
Okay, substituting we get, well I would tell you except I can't see the problem when I am typing. Hold on, I have to go out of this... Oh yeah, one of the points is (1,1), so let x = 1 when y = 1, and m = (2/3), and try to find b.
Last edited by John E. Franklin (20060510 14:13:30)
igloo myrtilles fourmis
Re: Algebra
problem # 1 continued...
Okay, so 1 for y = (2/3) times (1) for x plus b. Or y = mx + b, so 1 = (2/3)(1) + b. Multiply the 1 and add to other side. 1 + (2/3) = b Now you know b! Horray!!! And you know m! Ever noticed people avoid the ! mark because it means factorial in math?
igloo myrtilles fourmis
Re: Algebra
problem#1 continued... Simply put m and b into the y=mx+b equation, but leave the y and x as variables y and x, and this is your equation for the line! So ANSWER is: y = (2/3)x + 1 + (2/3), which is the same as y = 2x/3 + 5/3, and also same as y = (1/3)(2x + 5). Hope this helps a little bit. Also, you might have been taught a different way, so sorry if this isn't how you should do it.
igloo myrtilles fourmis
Re: Algebra
For the problem #3d, remember that if the point is on the line, then the equation will be a TRUE statement when you put in the x and the y numbers for that point into the equation the the line. Or another way to check might be to see if the slope from the point in question to one of the known points on the line is the slope of the line. See what I mean. If you know the slope of the line, which is the slant or direction the line points, then if the point you are trying to figure out is on the line, then you can subtract the y's with a point on the line you know and divide it by the subtraction of x's from wanttoknow point to known point and see if you get the m value of the known slope. Hope that helps.
igloo myrtilles fourmis
Re: Algebra
# 2
You're told that the line is parallel to 2x  5y = 6. This means that your line will have the same gradient as that line.
Rearranging to y = mx+b form gives y = 2/5x  6/5, so the gradient of your line is 2/5.
You are also told that your line passes through (2,3). Using this combined with the gradient, you can find the yintercept.
From the yaxis, the point has been moved left by 2, so you need to move it right 2 to get back there. You know that for every 1 that the point moves right, it goes 2/5 up, so you can work out that the yintercept is 3+(2*2/5) = 2 1/5, or 11/5.
Therefore, your line is 5y = 2x  11.
# 3a
You find slopes by dividing the ydifference by the xdifference. For PQ, that means (2(1)÷(41) = 3÷3 = 1
Similarly, for PR, the slope is (4(1)÷(61) = 5÷5 = 1.
# 3b
As the two slopes are the same, this means that all 3 points lie in a straight line. We could have also concluded this by looking at question 3c. =P
# 3c
We already know that the slope is 1, which means that x and y will always have the same difference. We can see that this is 2.
Therefore, y = x  2
# 3d
Using John's method, we can substitute the values in to see whether the equation is true or not. This gives 8 = 102.
The equation is true, and so the point lies on the line.
Why did the vector cross the road? It wanted to be normal.
