Answer:
Point-slope form: ![y-1 = \frac{1}{5}(x-10)](https://tex.z-dn.net/?f=y-1%20%3D%20%5Cfrac%7B1%7D%7B5%7D%28x-10%29)
Slope intercept form: ![y = 0.2x -1](https://tex.z-dn.net/?f=y%20%3D%200.2x%20-1)
Step-by-step explanation:
Point-slope form:
![y-y_1 = m(x-x_1)](https://tex.z-dn.net/?f=y-y_1%20%3D%20m%28x-x_1%29)
y1 and x1 are the y and x coordinates for a point on the line.
m is the slope.
We know m, as it is given:
m = 1/5
We also know a point on the line, (10, 1).
Here, the x-coordinate is 10, and the y-coordinate is 1.
Thus we can say the following:
![x_1 = 10\\y_1 = 1](https://tex.z-dn.net/?f=x_1%20%3D%2010%5C%5Cy_1%20%3D%201)
And, from previously:
![m=\frac{1}{5}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B1%7D%7B5%7D)
Let's put these values into our equation.
![y-y_1 = m(x-x_1)](https://tex.z-dn.net/?f=y-y_1%20%3D%20m%28x-x_1%29)
![y-1 = \frac{1}{5}(x-10)](https://tex.z-dn.net/?f=y-1%20%3D%20%5Cfrac%7B1%7D%7B5%7D%28x-10%29)
If we want to answer in slope-intercept form:
![y=ax+b](https://tex.z-dn.net/?f=y%3Dax%2Bb)
We can continue based on our point-form expression to isolate y:
![y-1 = \frac{1}{5}(x-10)\\](https://tex.z-dn.net/?f=y-1%20%3D%20%5Cfrac%7B1%7D%7B5%7D%28x-10%29%5C%5C)
Add 1 to both sides:
![y= \frac{1}{5}(x-10)+1](https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B1%7D%7B5%7D%28x-10%29%2B1)
Expanding the parenthesis (multiplying 1/5 with x and -10):
![y= \frac{1}{5}x -2 + 1 \\\\y= \frac{1}{5}x -1\\\\y = 0.2x -1](https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B1%7D%7B5%7Dx%20-2%20%2B%201%20%5C%5C%5C%5Cy%3D%20%5Cfrac%7B1%7D%7B5%7Dx%20-1%5C%5C%5C%5Cy%20%3D%200.2x%20-1)
Point-slope form: ![y-1 = \frac{1}{5}(x-10)](https://tex.z-dn.net/?f=y-1%20%3D%20%5Cfrac%7B1%7D%7B5%7D%28x-10%29)
Slope intercept form: ![y = 0.2x -1](https://tex.z-dn.net/?f=y%20%3D%200.2x%20-1)