1. ## Question about standard form of a line

Hi, I'm new here! I'm having an issue with this problem in my homework. I need to find the equation of each line and put it in standard form.

The points I have are (-7, -4) and (0,6).

So I used the y2-y1 method.

(6-4)
-----
(0-(-7) and got 10/7. So y= 10/7x + b
. Then I found b.

-4 = 10/7(-7) + b
-4 = 10 + b
-6 = b
so in slope-intercept form my answer is y= 10/7x - 6.

But i need it in standard form and the tutorials I've found for this say you just need to minus mx from both sides. So I did and I got -10/7x +y = -6. However my book says the answer should be 2x + 7y = -42.

I think I'm not understanding how to convert slope-intercept form to standard form. The other problems in my homework have the same issue.

By the way, I hope I put this in the correct format- sorry if I didn't! Thank you...

2. The standard form has integer coefficients.

3. Originally Posted by Nycea

The points I have are (-7, -4) and (0,6).

So I used the y2-y1 method.

(6-4)
-----
(0-(-7) and got 10/7. So y= 10/7x + b
. Then I found b.
[I][B]
-4 = 10/7(-7) + b
next step is $\displaystyle -4=-10+b \implies b = 6\neq -6$

4. Originally Posted by Nycea
The points I have are (-7, -4) and (0,6).
(6-4)
-----
(0-(-7) and got 10/7. So y= 10/7x + b
. Then I found b.

-4 = 10/7(-7) + b
-4 = 10 + b
-6 = b
so in slope-intercept form my answer is y= 10/7x - 6.
Yout slope is correct.
Your b is wrong: should be 6; 10/7(-7) = -10
Also quite evident from the given point (0,6)

5. I think I've got it! Thanks you guys! <3