Write an equation in standard form for a line passing through A(-2, 3) and B(3,4)

Question

I wanted to check whether the answer that I got was correct. The answer I believe it to be is x-y=1. Is this right?

Answer 1

You have provided an answer of x-y=1 and are asking if this is correct.

 

If this equation is correct, it will be true for both (x,y)=(-2,3) and (3,4).

x-y=1

For point A: (-2)-(3)=-2-3=-5, not 1, plugging the x and y value of point A into your equation

This, by itself, makes your equation incorrect because it has to be true for both A and B to be  correct.

For point B 3-4= -1, not 1. plugging the x and y value of B into your equation.

So your equation is not correct.

Note, again, that it was incorrect for both points; however, if it was true for one point and not the other, the equation would still not  be correct.

 

However, your equation is in standard form, Ax+By=C where A, B and C are integers with A not negative (A is positive or 0 and both B and C can be any integer, positive or negative or 0).

Careful.  A and B, in the definition of the standard form of a linear equation, are the coefficients of x an y (number multipliers of x and y) and NOT the points A and B of your problem.

 

In your case, A (the coefficient of x in the standard form equation) is understood to be 1 because x=1x, and B (the coefficient of y in the standard form equation) is understood to be -1 because -y=-1y.

 

Heads up.  Different instructors may allow A to be any integer as well as B and C.

Some instructors may require A, B and C to have no common factors other than 1.

For example: If A=2, B=6, and C=10, they all have the common factor 2 since A=1(2), B=3(2) and C=(5(2).  So your instructor may require you to "reduce"2x+6y=10 to x+3y=5 by dividing everything by the common factor, 2.  Also, some instructors may require A, B and C to not be zero so that x=5 would not be a standard form equation; and some instructors may allow A, B and C to be zero so that x=5 would be standard form because x=5 can be written as 1x+0y=5.

 

Now back to finding the correct equation.

You may be tempted to find the point slope form [y=mx+b] and convert it to standard form because that's what you are used to.

I would use point slope form [(y-y1)=m(x-x1)] to write an equation and convert it to standard form [Ax+By=C], because I think it is an easier process.

y-y1=m(x-x1)

where m is the slope and equals (y2-y1)/(x2-x1)

In your case, let point A= (x2,y2) and point B=(x1,y1), so x2=(-2), y2=3, x1=3 and y1=4

You could reverse wihich point is (x1,y1) and (x2,y2) if you wish.

m=(y2-y1)/(x2-x1)=(3-4)/(-2-3)=-1/-5=1/5

plug this m into y-y1=m(x-x1)

y-y1=(1/5)(x-x1)

You can now use either point, A or B, as (x1,y1).

For simplicity, let's keep with B=(x1,y1)=(3,4).

y-4=(1/5)(x-3)

Your temptation, at this point, would be to use the distributive property to "get rid of the paranthesis";

and that would not be wrong.  However, if you multiply both sides of the equation by the denominator of the slope, m, you will get to the answer quicker by immediately eliminating fractions.

5[y-4]=5(1/5)(x-3)

5y-20=1(x-3) using the distributive property on left of equals and 5(1/5)=1 on right.

5y-20=x-3 using the distributive property or the multiplicative identity on the right of equals.

-20+3=x-5y adding 3 and subtracting 5y on both sides of the equals.

-17=x-5y

x-5y=-17 using the property of equality that if a=b then b=a

To check or standard form equation, plug in A and B

-2-5(3)=-2-15=-17

3-5(4)=3-20=-17.

 

Now, for argument's sake, let's use point A for (x1, y1) in  y-y1=(1/5)(x-x1)

y-3=(1/5)(x-[-2])

5(y-3)=5(1/5)(x-[-2])

5y-15=x+2

-15-2=x-5y

-17=x-5y

x-5y=-17

So you can see it doesn't make any difference which point you use in y-y1=m(x-x1).