Answer:
The equation of a line with the slope of 1/4 that passes through the point (-8,2) is
![y=(x)/(4)+4](https://img.qammunity.org/2020/formulas/mathematics/middle-school/vti142rq780n4heoyf2pizgn9s6p67514e.png)
Solution:
Given, slope of a line is
and a point is (-8, 2)
We have to find the line equation with above given values.
Now, we know that, point slope form of a line is
![y-y_(1)=m\left(x-x_(1)\right)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ekhr1zq9rk6cqtf5v6gdiuqkcziiby04yk.png)
Where, m is slope of line and
![\left(x_(1), y_(1)\right) \text { is point on that line. }](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1prh0t9ze43n8pnbn336mymmyo8pq0obg4.png)
Here in our problem,
![\mathrm{m}=(1)/(4) \text { and }\left(\mathrm{x}_(1), \mathrm{y}_(1)\right)=(-8,2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/5t5ph6fu8rhikm6y6v3xbp0zv2gs3nqm4q.png)
So, substitute above values in general form.
![\begin{array}{l}{y-2=(1)/(4)(x-(-8))} \\\\ {y-2=(1)/(4)(x+8)}\end{array}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/522jeec5mrgo1zpnuevpveyrhh0p22g9re.png)
4(y – 2) = (x + 8)
4y – 8 = x + 8
x – 4y + 8 + 8 = 0
x – 4y + 16 = 0.
4y = x +16
![y=(x)/(4)+4](https://img.qammunity.org/2020/formulas/mathematics/middle-school/vti142rq780n4heoyf2pizgn9s6p67514e.png)
Hence the equation of a line with the slope of 1/4 that passes through the point (-8,2) is
![y=(x)/(4)+4](https://img.qammunity.org/2020/formulas/mathematics/middle-school/vti142rq780n4heoyf2pizgn9s6p67514e.png)