What is the equation of the line that passes through (-3,-1) and has a slope of 2/5? Put your answer in slope-intercept form.

Really need the help thx :)

Respuesta :

[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ -3}}\quad ,&{{ -1}})\quad \end{array} \\\quad \\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{2}{5} \\ \quad \\\\ % point-slope intercept y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad \textit{plug in the values and solve for "y"}\\ \left. \qquad \right. \uparrow\\ \textit{point-slope form}[/tex]

[tex]\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-(-1)=\cfrac{2}{5}(x-(-3)) \\\\\\ y+1=\cfrac{2}{5}(x+3)\implies y+1=\cfrac{2}{5}x+\cfrac{6}{5}\implies y=\cfrac{2}{5}x+\cfrac{6}{5}-1 \\\\\\ \begin{array}{llll} y=&\cfrac{2}{5}x&+\cfrac{1}{5}\\ &\uparrow &\quad \uparrow \\ &slope&y-intercept \end{array}\impliedby \textit{slope-intercept form}[/tex]

irspow
Slope-intercept form is y=mx+b where m=slope and b=y-intercept

We are told the slope is 2/5 so:

y=2x/5+b, using the point (-3-1) we can solve for b:

-1=2(-3)/5+b

-1=-6/5+b so

b=1/5 thus the line is:

y=(2x+1)/5