Respuesta :
Answer:
a. [tex]y=-x+-5[/tex]
b. [tex]y=\frac{4}{3}x+\frac{5}{3}[/tex]
Step-by-step explanation:
to solve part a we have the slope of the line and a point, so we use the point-slope equation
[tex]y=m(x-x_{1})+y_{1}[/tex]
where [tex]m[/tex] is the slope: [tex]m=-1[/tex], and [tex](x_{1},y_{1})[/tex] is the point, so in this case the point is [tex](0,-5)[/tex] thus [tex]x_{1}=0[/tex] and [tex]y_{1}=-5[/tex].Thus the equation is:
[tex]y=-1(x-0)+(-5)[/tex]
[tex]y=-x+-5[/tex]
And for part b we have two points. With the two points we can find the slope and then use the point-slope equation again.
To find the slope we use:
[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
for the points [tex](x_{1},y_{1})[/tex] and [tex](x_{2},y_{2})[/tex]. Since we have the poits: [tex](1,3)[/tex] and [tex](-2,-1)[/tex]
[tex]x_{1}=1\\y_{1}=3\\x_{2}=-2\\y_{2}=-1[/tex]
Thus, the slope:
[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-1-3}{-2-1}[/tex]
[tex]m=\frac{-4}{-3}[/tex]
[tex]m=\frac{4}{3}[/tex]
and now using the point-slope equation:
[tex]y=m(x-x_{1})+y_{1}[/tex]
[tex]y=\frac{4}{3}(x-1)+3\\y=\frac{4}{3}x-\frac{4}{3} +3\\y=\frac{4}{3}x+\frac{5}{3}[/tex]
