Question 256375
solve for the slope intercept form of the first equation of:


2y + x + 3 = 0. 


subtract x and subtract 3 from both sides of the equation to get:


2y = -x - 3


divide both sides of the equation by 2 to get:


y = (-1/2)*x - (3/2)


solve for the slope intercept form of the second equation of:


ax + 3y + 2 = 0.


subtract ax and subtract + 2 from both sides of the equation to get:


3y = -ax - 2


divide both sides of the equation by 3 to get:


y = (-a/3)*x - (2/3)


slope intercept form of these equations is y = m*x + b where m is the slope and b is the y-intercept.


your 2 equations in slope intercept form are:


y = (-1/2)*x - (3/2)
y = (-a/3)*x - (2/3)


if the graphs of these equations intersect at right angles, this means the lines formed by these equations are perpendicular to each other.


if the lines are perpendicular to each other, then their slopes are negative reciprocals of each other.


this means that the slope of one of the lines is equal to the negative reciprocal of the slope of the other line.


this means that:


(-1/2) = -1 / (-a/3)


multiply both sides of this equation by (-a/3) to get:


(-1/2)*(-a/3) = -1


multiply both sides of this equation by 6 to get:


(6)*(-1/2)*(-a/3) = -6


this simplifies to:


a = -6


the lines formed by these equations should be perpendicular when a = 6


take your second equation of:


y = (-a/3)*x - (2/3) and replace a with -6 to get:


y = (-(-6/3)*x) - (2/3) which becomes:


y = (6/3)*x - (2/3) which becomes:


y = 2*x - (2/3).


negative reciprocal of 2 is -(1/2) so the lines are perpendecular.


the two equation are:


y = (-1/2)*x - (3/2)
y = (2)*x - (2/3)


this happens when a = -6.


graph of these 2 equations is shown below:


{{{graph(600,600,-5,5,-5,5,(-1/2)*x - (3/2),2*x - (2/3))}}}