okay so im doing linear equations and these set of problems says standard form. so i know what it is: Ax+By=C but i dont understand how to get it or what you do. can you help me ive tried and im so confused! Thanks so much:)
so here are some of the problems:
1. (-4,3) and (-5,8)
2. (2,5) and (1,3)
3. (0,0) and (-2,4)
4. (12,-15) and (10,-6)
5. (5,-1) and (8,-3)
6. (10,-1) and (10,-7)
THANK YOU SO MUCH!!!!!!!
i need help really bad.
Standard form math help! linear equations.. please?
There is a form called slope-intercept form. The slope intercept form gives the equation of a line in terms of two points and a slope. Here it is:
(y2-y1)=m(x2-x1)
It comes from the definition of slope (rise over run or change in y over change in x). (x1,y1) is one point and (x2,y2) is another point. m is slope. To find the equation from two points, plug in the points and find the slope. Then put x and y back in (see below).
1) (-4,3) %26amp; (-5,8)
(8-3)=m(-5+4)
5=-m
m=-5
(y-3)=-5(x+4)
y-3=-5x-20
y+5x-3=-20
y+5x=-17
5x+y=-17
This is standard form. If you end up with a fractional coefficient, remember to multiply by the denominator. Without further ado, these are the remaining answers:
1) 5x+y=-17
2) -2x+y=1
3) 2x+y=0
4) 9x+2y=78
5) 2x+3y=7
6) x=10
On this one, the same x value results in two y values. This must mean that the line is vertical. Since both points are 10 units from the y-axis, the equation must be x=10.
Note: You can generalize the formula and write a program to do this for you! If you work the algebra correctly, you end up with the formula:
Y - [(y2-y1)/(x2-x1)]*X = [(y2-y1)/(x2-x1)]*x1-y1
If you have a TI-83/84+, then you can create a program ([PRGM][%26gt;][%26gt;][ENTER]). Name it whatever you want and press enter. Then type:
[PRGM][%26gt;][8][ENTER]
[PRGM][%26gt;][ENTER][ALPHA][+][X][1][2nd][...
[PRGM][%26gt;][ENTER][ALPHA][+][APHA][1][1]...
[PRGM][%26gt;][ENTER][ALPHA][+][X][2][2nd][...
[PRGM][%26gt;][ENTER][ALPHA][+][APHA][1][2]...
You should end up with:
ClrHome
Input "X1=",A
Input "Y1=",B
Input "X2=",C
Input "Y2=",D
Add the follow to the end:
{-(D-B)/(C-A),1,-(D-B)/(C-A)*A+B}
After you are done, quit [2nd][MODE] and press [PRGM] and select the program you created. This makes it so much easier! Be warned, this program will give decimal answers, so if there is a decimal, multiply by something reasonable to get it to all whole numbers (i.e. if you end up with {0.666666667,1,0.3333333333}, then type in *3[ENTER] and you will get whole number coefficients. By the way, the program returns {A,B,C} so your answer will be Ax+By=C.
Best of luck.
Reply:Are you trying to graph a line containing the points? Sets of coordinates don't tell you enough to do the problems... what are you attempting to do?
Reply:I'll do number 1 so you see how it's done. You just find the equation in slope-intercept form and then convert it to standard form.
(-4, 3) and (-5, 8)
find the slope using:
m = (y2 - y1)/(x2 - x1)
m = (8 - 3)/(-5 - -4)
m = -5
then put this into the form
y = mx + b
you now know m, so sub it in
y = -5x + b
to find b sub in one of the points, I'll sub in (-4, 3)
3 = -5(-4) +b
3 = 20 + b
b = -17
so y = -5x - 17
you can sub in the other point to check if you want
then convert this form to standard form
y = -5x - 17
5x + y = -17
Hope all that helps :)
Reply:i dont know how to do standard form with these, but you can use point slope (y=mx+b) ... you have to find the slope, for example, on number one, this is how you would find the slope:
-5- -4= -1
8-3=5
-1/5 = slope
then, you plot the points you have,connect them, and extend the line if needed so that you can find the y intercept (the point where the line hits the y- axis)
and then you plug your data into the equation:
y= -1/5 x + -16
simplified: y= -1/5x-16
(im not sure if the y intercept is 16, it was according to my HAND DRAWN graph. lol)
hope i helped at least a little!! :D
hollyhock
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