How do I write a General Linear Equation(Ax+By=C) through the 2 points(0,0) & (2,3)?

I don't know if this is how it's supposed to be done, but it's how I do it

If you transform that equation into y=mx+b form, you will find that m=a/-b. And you can find m by taking (y1 - y2)/(x1 - x2)

m = (0 - 3) / (0 - 2)

m = -3 / -2 = 3 / 2

a/-b = 3/2

a = 3

-b = 2

b = -2

3x - 2y = C

Just use one of the points to solve for C

3(0) - 2(0) = C

0 - 0 = C

0 = C

3x - 2y = 0

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in case you ever took precalculus or another math type which covers properties of purposes, you how you could nevertheless make certain the area of the function. between the purple flags to look out for is once you divide via 0. on your expression, while x = 3, the denominator is 0. for this reason, the area is each huge type beside 3. regrettably, the huge type you attempt to plug in (x = 3) is the only huge type that would not paintings in this function. to make certain this popular hand, you0 can graph this function on a TI-eighty 3. in case you zoom in near to the graph at x = 3, you will see that there is a sparkling spot there! this is because of the fact, as reported above, there in basic terms isn't a value of the expression at x = 3. you are going to be able to assert, nicely curiously like the respond could desire to be 6, observing the graph. this concept of what the respond "could desire to be" is what limits are all approximately. The values of the function on the left and magnificent of x = 3 all bypass in the direction of 6 as you get closer and closer. So we are saying the cut back as x is going to 3 is 6. So in spite of the undeniable fact that that's no longer technically the respond, 6 is your ideal selection. 0 will by no potential be suitable in any experience. the very ideal answer is to assert that the expression is undefined at x = 3. This challenge illustrates why 0/0 is stated as indeterminate. in this challenge, 0/0 in a manner equals 6. the concept 0/0 can equivalent something is easily the essence of calculus.

Since you have 2 points you can find the slope of the line, and using one of those points you can put the line's equation into point slope form which can then be manipulated into Standard Form...

m = (y2-y1) / (x2-x1) = (2-0) / (3-0) = 2/3

y - y1 = m (x - x1) Point Slope Form use the ordered pair (2.3)

y - 3 = 2/3 ( x -2) just simplify

y - 3 = 2/3x - 4/3

y = 2/3x + 5/3

-2/3x + y = 5/3 you can easily get rid of the troublesome fraction and

negative by multiplying through by -3

2x - 3y = -5

First, substitute the points into the equation:

0a+0b=c

2a+3b=c

Note that first equation reduces (most won't)

c=0

2a+3b=0

Usually, you'll have 2 equations and 3 unknowns (a,b,c)... you'll have to pick one of the numbers yourself. For example, all these equations are equivalent:

a+b=1

2a+2b=2

3a+3b=3

But I digress... 2a+3b=0... pick one, and make it easy... a=1

3b+2=0

Solve... usually it won't be this easy, but there are a lot of ways to solve a 2-variable system and I'm running out of room

b=-2/3

so you have a solution:

1x-2/3y=0

Simplify by multiplying out the fraction:

3x-2y=0 <-- ANSWER

Wait... did I just do your homework for you?