Question

Find the equation of the line that contains the point (-1, -4) and is parallel to the line 6x+7y=13 . Write the equation in​ slope-intercept form, if possible.

Answer 1

Answer:  $y=-\dfrac67x-\dfrac{34}{7}$

Step-by-step explanation:

Slope-intercept form: y=mx+c   (i) , of linear equations,  m= slope and c= y-intercept of the line.

6x+7y=13in slope intercept form would be :

$7y=13-6x\\\\\Rightarrow\ y=\dfrac{-6}{7}x+\dfrac{13}{7}$

Comparing to (i) , $m = \dfrac{-6}{7}$

Slope of parallel lines are equal.

So, slope of required line = $m = \dfrac{-6}{7}$

Equation of line with slope m and passes through point : $(y-b)=m(x-a)$

So, equation of required line:

$(y-(-4))=\dfrac{-6}{7}(x-(-1))\\\\\Rightarrow\ (y+4)=\dfrac{-6}{7}(x+1)\\\\\Rightarrow\ y+4=\dfrac{-6}{7}x-\dfrac{6}{7}\\\\\Rightarrow\ y=-\dfrac67x-\dfrac{6}{7}-4\\\\\Rightarrow\ y=-\dfrac67x-\dfrac{34}{7}$