anonymous one year ago The point-slope form of the equation of the line that passes through (–4, –3) and (12, 1) is y – 1 = (x – 12). What is the standard form of the equation for this line? x – 4y = 8 x – 4y = 2 4x – y = 8 4x – y = 2

1. anonymous

wow another set of wrong answers?

2. anonymous

Standard form is ax + by = c

3. anonymous

they gave it to me this way how are they wrong

4. anonymous

the slope of the line through $$(-4,-3)$$ and $$(12,1)$$ is $\frac{1-(-3)}{12-(-4)}=\frac{4}{16}=\frac{1}{4}$

5. anonymous

so the point slope form is not $y-1=(x-12)$ it is $y-1=\frac{1}{4}(x-12)$

6. anonymous

7. anonymous

can u help me with two more

8. anonymous

$x-4y=8$ is correct

9. anonymous

sure

10. anonymous

The point-slope form of the equation of the line that passes through (–5, –1) and (10, –7) is y + 7 = (x – 10). What is the standard form of the equation for this line? 2x – 5y = –15 2x – 5y = –17 2x + 5y = –15 2x + 5y = –17

11. anonymous

wow still wrong but you do have a correct answer there

12. anonymous

13. anonymous

14. anonymous

go with C

15. anonymous

Line CD passes through points C(1, 3) and D(4, –3). If the equation of the line is written in slope-intercept form, y = mx + b, what is the value of b? –5 –2 1 5

16. anonymous

slope is $\frac{-3-3}{4-1}=\frac{-6}{3}=-2$

17. anonymous

aye i got 1 more under it

18. anonymous

point slope gives $y-3=-2(x-1)$solve for $$y$$ via $y-3=-2x+2\\ y=-2x+5$ so $$b=5$$