Question 922400
For what value(s) of k will the lines
kx+5y=2 and 2x+(k+3)y=4 be: a) parallel to each other?
:
put both equations in the slope intercept form
kx + 5y = 2
5y = -kx + 2
y = {{{-k/5}}}x + {{{2/5}}}
and
2x + (k+3)y = 4
(k+3)y = -2x + 4
y = {{{-2/((k+3))}}}x + {{{4/((k+3))}}}
the slope of parallel lines are equal; therefore\
{{{-k/5}}} = {{{-2/((K+3))}}}
cross multiply
-k(k+3) = 5 * -2
-k(k+3) = -10
multiply both sides by -1
k(k+3) = 10
k^2 + 3k - 10 = 0
factors to
(k+5)(k-2) = 0
the positive solution
k = 2
the negative solution will also work
k = -5
:
:
b)perpendicular to each other
the relationship of slopes of perpendicular lines: m1 * m2 = -1, therefore
{{{-k/5}}} * {{{-2/((k+3))}}} = -1
{{{(2k)/(5(k+3))}}} = -1
{{{(2k)/((5k+15))}}} = -1
2k = -1(5k+15)
2k = -5k - 15
2k + 5k = -15
7k = -15
k = {{{-15/7}}} value for perpendicular lines