Question 113488


First find the slope for the line {{{3x - 6y=0}}}

the slope is:

{{{3x - 6y=0}}}…..solve for {{{y}}}; move {{{3x}}} to the right 

{{{- 6y=-3x}}}….. divide both sides by {{{-6}}}
 
{{{- 6y/-6=-3x/-6}}}…..

{{{y=(1/2)x}}}…..

=>…{{{m=1/2}}}

Since the line which we are looking for is perpendicular to the line 

{{{3x - 6y=0}}}, and we know that {{{perpendicular}}}{{{ lines}}} have 

{{{negative}}}{{{ reciprocal}}}{{{ slopes}}}, then the slope of the line which 

we are looking for is:

{{{m1 = - 1/m}}}

{{{m1 = - 1/(1/2)}}}

{{{m1 = -2}}}….this is a slope of the line which we are looking for


We also know that the line which we are looking for passing through (-2,-1)

Now, we can find equation of a line {{{by }}}{{{slope}}} and {{{one}}}{{{ point}}}. Here is solution:


*[invoke find_line_by_slope_and_point -2, -1, -2]


the equation in standard form is:

{{{2x + y + 5 = 0}}}