Write the line parallel to the line with the equation x-5y=6 and through point (10,2) in standard form.
If anyone can explain this to me, i would greatly appreciate it.
Standard form for a line is $\displaystyle Ax+By=c$ where $\displaystyle \frac{-A}{B}$ is the slope of the line. So in order for the line you want to be parallel to x-5y=6, it needs to have the same slope. Solving this equation by subracting x from both sides and then dividing by -5 we get $\displaystyle x-5y-x=6-x=-5y=6-x$ so $\displaystyle \frac{-5y}{-5}=\frac{6-x}{-5}$ so $\displaystyle y=\frac{6}{-5}+\frac{x}{5}$ so the slope is $\displaystyle \frac{1}{5}$
So our line needs to be y=mx+b and we now know $\displaystyle m=\frac{1}{5}$
now we need it to go through the point (10,2), so we can plug that point into the equation to solve for b. 10 is x and 2 is y so $\displaystyle 2=\frac{1}{5}(10)+b$ and solving for b, $\displaystyle b=2-2=0$
So $\displaystyle y=\frac{1}{5}x$ which is parallel to the given line and contains the point (10,2)
To get it in standard form simply subtract $\displaystyle \frac{1}{5}x$ from both sides to get $\displaystyle \frac{-1}{5}x+y=0$