Respuesta :
[tex]y+4=-\frac{1}{8} \times(x+6)[/tex] or [tex]y+5=-\frac{1}{8} \times(x-2)[/tex] is the slope point form of the line.
Step-by-step explanation:
The given points are (-6,-4) and (2,-5) are as [tex]\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right)[/tex]. Need to find out the point slope form of the line which passes through these points. The formula of slope point form for an equation of a straight line as
[tex]\left(y-y_{1}\right)=\text { slope } \times\left(x-x_{1}\right)[/tex]
Hence, we have two points and we have to find out first slope of the line
[tex]\left(y-y_{1}\right)=m \times\left(x-x_{1}\right)[/tex]
Where m is the slope and [tex]\left(x_{1}, y_{1}\right)[/tex] is a point the line passes through. Hence, we have two points and we have to find out first slope of the line
[tex]\text {slope}=\frac{-5-(-4)}{2-(-6)}=-\frac{1}{8}[/tex]
Now, slope point form of the line is,
[tex]y-(-4)=-\frac{1}{8} \times(x-(-6))[/tex]
[tex]y+4=-\frac{1}{8} \times(x+6)[/tex]
Similarly at (2,5),
[tex]y-(-5)=-\frac{1}{8} \times(x-2)[/tex]
[tex]y+5=-\frac{1}{8} \times(x-2)[/tex]
