or
is the slope point form of the line.
<u>Step-by-step explanation:</u>
The given points are (-6,-4) and (2,-5) are as
. 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
![\left(y-y_{1}\right)=\text { slope } \times\left(x-x_{1}\right)](https://tex.z-dn.net/?f=%5Cleft%28y-y_%7B1%7D%5Cright%29%3D%5Ctext%20%7B%20slope%20%7D%20%5Ctimes%5Cleft%28x-x_%7B1%7D%5Cright%29)
Hence, we have two points and we have to find out first slope of the line
![\left(y-y_{1}\right)=m \times\left(x-x_{1}\right)](https://tex.z-dn.net/?f=%5Cleft%28y-y_%7B1%7D%5Cright%29%3Dm%20%5Ctimes%5Cleft%28x-x_%7B1%7D%5Cright%29)
Where m is the slope and
is a point the line passes through. Hence, we have two points and we have to find out first slope of the line
![\text {slope}=\frac{-5-(-4)}{2-(-6)}=-\frac{1}{8}](https://tex.z-dn.net/?f=%5Ctext%20%7Bslope%7D%3D%5Cfrac%7B-5-%28-4%29%7D%7B2-%28-6%29%7D%3D-%5Cfrac%7B1%7D%7B8%7D)
Now, slope point form of the line is,
![y-(-4)=-\frac{1}{8} \times(x-(-6))](https://tex.z-dn.net/?f=y-%28-4%29%3D-%5Cfrac%7B1%7D%7B8%7D%20%5Ctimes%28x-%28-6%29%29)
![y+4=-\frac{1}{8} \times(x+6)](https://tex.z-dn.net/?f=y%2B4%3D-%5Cfrac%7B1%7D%7B8%7D%20%5Ctimes%28x%2B6%29)
Similarly at (2,5),
![y-(-5)=-\frac{1}{8} \times(x-2)](https://tex.z-dn.net/?f=y-%28-5%29%3D-%5Cfrac%7B1%7D%7B8%7D%20%5Ctimes%28x-2%29)
![y+5=-\frac{1}{8} \times(x-2)](https://tex.z-dn.net/?f=y%2B5%3D-%5Cfrac%7B1%7D%7B8%7D%20%5Ctimes%28x-2%29)