Question

Find an equation of the line through these points (15,2.2) (5,1.6). Write answer in a slope-intercept form

Answer 1

Answer:

$y=\frac{\displaystyle 3}{\displaystyle 50}x+\frac{\displaystyle 13}{\displaystyle 10}$

Step-by-step explanation:

Hi there!

Slope-intercept form: $y=mx+b$ where $m$ is the slope and $b$ is the y-intercept (the value of y when x is 0)

1) Determine the slope (m)

$m=\frac{\displaystyle y_2-y_1}{\displaystyle x_2-x_1}$ where two given points are $(x_1,y_1)$ and $(x_2,y_2)$

Plug in the given points (15,2.2) and (5,1.6):

$m=\frac{\displaystyle 1.6-2.2}{\displaystyle 5-15}\\\\m=\frac{\displaystyle -0.6}{\displaystyle -10}\\\\m=\frac{\displaystyle 0.6}{\displaystyle 10}\\\\m=\frac{\displaystyle 0.3}{\displaystyle 5}\\\\m=\frac{\displaystyle 3}{\displaystyle 50}$

Therefore, the slope of the line is $\frac{\displaystyle 3}{\displaystyle 50}$. Plug this into $y=mx+b$:

$y=\frac{\displaystyle 3}{\displaystyle 50}x+b$

2) Determine the y-intercept (b)

$y=\frac{\displaystyle 3}{\displaystyle 50}x+b$

Plug in a given point and solve for b:

$1.6=\frac{\displaystyle 3}{\displaystyle 50}(5)+b\\\\1.6=\frac{\displaystyle 3}{\displaystyle 10}+b\\\\1.6-\frac{\displaystyle 3}{\displaystyle 10}=\frac{\displaystyle 3}{\displaystyle 10}+b-\frac{\displaystyle 3}{\displaystyle 10}\\\\\frac{\displaystyle 13}{\displaystyle 10}=b$

Therefore, the y-intercept is $\frac{\displaystyle 13}{\displaystyle 10}$. Plug this back into $y=\frac{\displaystyle 3}{\displaystyle 50}x+b$:

$y=\frac{\displaystyle 3}{\displaystyle 50}x+\frac{\displaystyle 13}{\displaystyle 10}$

I hope this helps!