Question

Find the equation of the line that contains the point (-1,-11) and is parallel to the line 7x+3y=10

Answer 1

Y=- \frac{7}{3} $-13 \frac{1}{3}$.

To find the equation of a line, you need two things: the slope and the y-intercept. 

The slopes of parallel lines are the same. So we can find the slope of the new line by finding the slope of the first line. To do that, we need to put it in y=mx+b format, where m is the slope. So we must rearrange the 7x+3y=10. First subtract 7x from both sides to make it look like:
       $3y=10-7x$
Then divide both sides three:
       b$y= \frac{10}{3} - \frac{7}{3} x$
So now that it's in y=mx+b format, we can now see that the m= - \frac{7}{3}

Now we know the m of the new equation, we need to find the b, or the y-intercept. To do this, we can plug the point we have and the m value into the y=mx+b format.
       $(-11)=- \frac{-7}{3} (-1) + b$
Solving this, we can subtract 7/3 from both sides:
     $-11- \frac{7}{3} = b$
Therefore, b= $-13 \frac{1}{3}$

Plugging the m= - \frac{7}{3} and the b= $-13 \frac{1}{3}$ back into the y=mx+b format, your parallel line is y=- \frac{7}{3} $-13 \frac{1}{3}$.