Answer:
[tex]y=-\frac{5}{3}x-\frac{11}{3}[/tex]
Step-by-step explanation:
We need to write the equation in slope-intercept form which passes through points (1/5,-4) and (16/5,-9)
Slope is given formula:
[tex]m=\frac{y_2-y_1}{x_2-x_1}=\frac{-9-\left(-4\right)}{\frac{16}{5}-\frac{1}{5}}=\frac{-9+4}{\frac{15}{5}}=\frac{-5}{3}[/tex]
Now plug the value of slope m and any one of the given point into point slope formula:
[tex]y-y_1=m\left(x-x_1\right)[/tex]
[tex]y-\left(-4\right)=\frac{-5}{3}\left(x-\frac{1}{5}\right)[/tex]
[tex]y+4=-\frac{5}{3}x+\frac{1}{3}[/tex]
[tex]y=-\frac{5}{3}x+\frac{1}{3}-4[/tex]
Now simplify this and write in slope intercept form y=mx=b
[tex]y=-\frac{5}{3}x-\frac{11}{3}[/tex]
Hence final answer is [tex]y=-\frac{5}{3}x-\frac{11}{3}[/tex]
Answer:
y=[tex]\frac{-5}{3}[/tex]x[tex]\frac{-11}{3}[/tex]
Step-by-step explanation:
y'-y1=m(x-x1)...........(1)
given
(x1,y1)=([tex]\frac{1}{5}[/tex],-4)
(x2,y2)=([tex]\frac{16}{5}[/tex],-9)
we know that slope=m=([tex]\frac{y2-y1}{x2-x1}[/tex])
put values of (x1,y1) and (x2,y2)in above formula
we get
m=([tex]\frac{y2-y1}{x2-x1}[/tex])
m=([tex]\frac{-9-(-4)}{16/5-1/5}[/tex])
m=([tex]\frac{-9-+4)}{15/5}[/tex])
m=([tex]\frac{-5)}{15/5}[/tex])
m=([tex]\frac{-5)}{3}[/tex])
now put slope (m) and points (x1,y1) in (1)
y-y1=m(x-x1)
y-(-4)=-5/3(x-1/5)
y+4= (-5/3)x+1/3
finally we get
y=[tex]\frac{'-5}{3}[/tex]x+[tex]\frac{1}{3}[/tex]-4
y=[tex]\frac{-5}{3}[/tex]x[tex]\frac{-11}{3}[/tex]
