Question

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

Answer 1

wow another set of wrong answers?

Answer 2

Standard form is ax + by = c

Answer 3

they gave it to me this way how are they wrong

Answer 4

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

Answer 5

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

Answer 6

first answer is however, correct

Answer 7

can u help me with two more

Answer 8

\[x-4y=8\] is correct

Answer 9

sure

Answer 10

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

Answer 11

wow still wrong
but you do have a correct answer there

Answer 12

you need method or answer?

Answer 13

answer

Answer 14

go with C

Answer 15

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

Answer 16

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

Answer 17

aye i got 1 more under it

Answer 18

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