Question

What is an equation of a line, in point-slope form, that passes through (1, – 7) and has a slope of -2/3?A. y - 7 = 2\3 (x-1)B. y + 7 = -2\3 (x-1)C. y - 7 = -2\3 (x+1)D. y + 7 = 2\3 (x+1)

Answer 1

the general equation of the line is

$y=mx+b$

where m si the slope and b an initial point , we have the slope and a point (x,y) to replace on the equation an calculate b

so

$\begin{gathered} (-7)=-\frac{2}{3}(1)+b \\ \\ -7=-\frac{2}{3}+b \\ b=-7+\frac{2}{3} \\ b=\frac{-19}{3} \end{gathered}$

then replace m and b on the general equation and we have the solution

$y=-\frac{2}{3}x-\frac{19}{3}$

this is the equation but the option have another form

on this case you can replace the point (1,-7) and check the solution, to rule out some options we know that the slope must be -2/3

so A isnt because the slope is 2/3

now check B

$\begin{gathered} y+7=-\frac{2}{3}(x-1) \\ (-7)+7=-\frac{2}{3}((1)-1) \\ 0=-\frac{2}{3}(0) \\ 0=0 \end{gathered}$

equality is correct

so the answer is B

we can check C because the slope is -2/3

$\begin{gathered} y-7=-\frac{2}{3}(x+1) \\ (-7)-7=-\frac{2}{3}((1)+1) \\ -14=-\frac{2}{3}(2) \\ -14=-\frac{4}{3} \end{gathered}$

the equality is wrong

this confirms that the answer is B