What is the equation of the line with slope  m= -6  that passes through  (-11,3) ?

2 Answers
Jan 28, 2016

$y = - 6 x - 63$

Explanation:

The standard equation of a line is $y = m x + c$, so we get $y = - 6 x + c$.

Now, since the line passes through the point, the point has to satisfy the equation of the line. Substitute $\left(- 11 , 3\right)$ in the equation to get:
$3 = - 6 \left(- 11\right) + c \implies c = - 63$.

Thus, the equation of the line becomes $y = - 6 x - 63$.

Jan 28, 2016

$6 x + y + 63 = 0$

Explanation:

SUPPOSE, THE EQUATION OF THE STRAIGHT LINE IS,
$y = m x + c$
where $c$ is unknown.
now, in the problem,
$m = - 6$
and the line goes through $\left(- 11 , 3\right)$ point.
now, by passing the equation of the straight line through $\left(- 11 , 3\right)$ point and putting $m = - 6$ in the equation, we get,
$3 = \left(- 6\right) \left(- 11\right) + c$
$\mathmr{and} , 3 = 66 + c$
$\mathmr{and} , c = 3 - 66$
$\mathmr{and} , c = - 63$
now, by putting $m = - 6$ and $c = - 63$ in the first equation, we will get the equation of the straight line.
so, the equation of the straight line is,
$y = - 6 x - 63$
$\mathmr{and} , 6 x + y + 63 = 0$