Question 644646
4.  through ({{{-8}}},{{{8}}}) and {{{parallel}}} to {{{y = (3/2)x -3}}}.  

*[invoke equation_parallel_or_perpendicular "parallel", 3/2, -3, -8, 8]



5.  through ({{{-3}}},{{{8}}}) and {{{vertical}}}... I guess to line {{{y = (3/2)x -3}}}

 *[invoke equation_parallel_or_perpendicular "perpendicular", 3/2, -3, -3, 8]


6.  through ({{{-7}}},{{{5}}}) and perpendicular to {{{y=(2/3)x -2}}}

*[invoke equation_parallel_or_perpendicular "perpendicular", 2/3, -2, -7, 5]

 I think you can't see everything here and I will write remaining part:

So the perpendicular slope is {{{m=-3/2}}}

So now we know the slope of the unknown line is {{{m=-3/2}}} (its the negative reciprocal of {{{2/3}}} from the line {{{y=(2/3)x-2}}}. Also since the unknown line goes through ({{{-7}}},{{{5}}}), we can find the equation by plugging in this info into the point-slope formula

Point-Slope Formula:

{{{y-y1=m(x-x)}}} where m is the slope and ({{{x1}}},{{{y1}}}) is the given point


{{{y-5=-(3/2)(x+7)}}}

{{{y-5=-(3/2)x+7(-3/2)}}}

{{{y-5=-(3/2)x-21/2}}}

{{{y=-(3/2)x-21/2+5}}}

{{{y=-(3/2)x-21/2+10/2}}}

{{{y=-(3/2)x-11/2}}}

So the equation of the line that is perpendicular to {{{y=(2/3)x-2}}} and goes through ({{{-7}}},{{{5}}}) is {{{y=-(3/2)x-11/2}}}



{{{ graph( 600, 600, -10, 10, -10, 10, (2/3)x-2, -(3/2)x-11/2) }}}