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

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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

Mathematics
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wow another set of wrong answers?
Standard form is ax + by = c
they gave it to me this way how are they wrong

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Other answers:

the slope of the line through \((-4,-3)\) and \((12,1)\) is \[\frac{1-(-3)}{12-(-4)}=\frac{4}{16}=\frac{1}{4}\]
so the point slope form is not \[y-1=(x-12)\] it is \[y-1=\frac{1}{4}(x-12)\]
first answer is however, correct
can u help me with two more
\[x-4y=8\] is correct
sure
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
wow still wrong but you do have a correct answer there
you need method or answer?
answer
go with C
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
slope is \[\frac{-3-3}{4-1}=\frac{-6}{3}=-2\]
aye i got 1 more under it
point slope gives \[y-3=-2(x-1)\]solve for \(y\) via \[y-3=-2x+2\\ y=-2x+5\] so \(b=5\)

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