How do you solve the equation of each line in standard form if i am only given the x-intercept and the y-intercept?

Answer 1

Answer: Given y-intercept b and x-intercept x2 , the equation of the line in standard form will be: bx+x2y-bx2 =0

Explanation one

First off, the standard form of an equation is Ax +By = C.

If you have both intercepts, then you have two points. A line can easily be defined by two points. If we let the y intercept be y1 i.e the point P1 =(0,y1) and the x intercept be x2 i.e. the point P2 =(x2,0). Then we can use those two points to calculate slope:

m= (y1-0)/(0-x2)= -(y1/x2).

Now, we can use the point slope formula.

That formula is (y-y1)=m(x-x1 ) where (x1 ,y1 ) is any given point. ( NOT a variable, but a specific point)

Now since we have two points P1 and P2 we can use either.

Let's use P2

(y-0)=m(x-x2) Simplify:

y = mx - mx2

Now note that (mx2 ) is just a number, not a variable. It is the slope multiplied by the x-intercept. We can call mx2 =C

So lastly, we put the equation in standard from.

Ax+By+C=0

-mx+y+mx2 =0

We are not quite done. Remember we know exactly what m is.

So let's substitute m=-(y1/x2) into the equation.

-(-(y1/x2))x+y-(y1/x2)x2 =0

(y1 /x2 )x+y+(-y1)=0

or

(y1 /x2 )x+y-y1=0

We could also be rid of the x2 in the denominator and make it look much nicer.

y1x+x2y-x2y1 =0

If you prefer, since the y intercept is often denoted by b, we could let y1 =b and we have

bx+x2y-x2 b =0

(we could take this one step further, let's call the x intercept t

then given b and t we could write the equation in standard form as

bx+ty-tb=0)

so A=y1

B=x2

C=-x2y1

For instance. if you had P2 =(1,0) and P1=(0,3), you can plug in the values directly from the formula we just calculated. A=3, B=1 and C=-3

3x+y-3=0

If we put this in slope intercept form, we have

y=-3x+3 which tells us right away the slope is -3 and the y intercept is 3.

If we calculated slope with P2 =(1,0) and P1=(0,3) we would have m=3/-1=-3

and y intercept is 3 since we have the point P1=(0,3)

So we could write that equation in slope intercept from right away as y=-3x+3

or in standard form as 3x+y-3=0

Explanation two

There is a form of the line know as two point form.

Given points P1 =(x1 ,y1 ) and P2 =(x2 ,y2 )

The equation of the line though those two points is:



(y-y1)/(y2 -y1 )=(x-x1)/(x2 -x1)

If we now plug in P1 =(0,y1) and P2 =(x2,0)

we have

(y-y1)/(0 - y1)=(x-0)/(x2)=(y- y1)/ -y1=x/x2 or -(y-y1)/ y1=x/x2

Now let's cross multiply so have y1 x=-(y-y1)x2 or after simplifying.

y1 x= -yx2+y1x2

y1 x + x2 y - y1x2=0 which agrees with our formula above. If we let b=y1 we have

b x + x2 y- bx2=0

Technical note: The y intercept is the y-value of the point where the graph of a function or relation intercepts the y-axis of the coordinate system. So if we have y=mx+b, the y intercept is b, and the point (0,b) is on the line.

Explanation 3. using a numerical example.

Write the information as two ordered pairs like: (0,5) y intercept and (2,0) x intercept.

Then use the method you are most familiar with to write the equation.

Using the slope formula: m = change in y/ change in x = 0-5/2 -0 = -5/2 is slope.

Replace m and b in the slope intercept form so the answer is:

y = -5/2 x + 5. This is the equation of a line with intercepts (0,5) and (2,0).

To change into Standard form we multiply by 2 to eliminate the fractions so

2y = -5x + 10 and then add the 5x term to both sides giving: 5x + 2y =10.