Question 83173
#38

{{{drawing( 300, 300, -7, 7, -7, 7,
  rectangle( -2, -1, 4, 5 ),
  line(-2, -1, 4, 5 ),
  line(-2,5,4,-1),
  circle(-2,-1,.1),
  circle(-2,-1,.15),
  circle(-2,5,.1),
  circle(-2,5,.15),
  circle(4,5,.1),
  circle(4,5,.15),
  circle(4,-1,.1),
  circle(4,-1,.15),
  locate( -2.5, -1.25, "(-2,-1)" ),
  locate( -2.5, 5.75, "(-2, 5)" ),
  locate( 4, -1, "(4, -1)" ),
  locate( 4, 5, "(4, 5)" )

)}}}


If we construct the figure that the problem is describing, we get a square with the given vertices and 2 lines. We need to find the slopes of the 2 lines to see if they are perpendicular. So the first line goes through the points (-2,5) and (4,-1) (it is the line that is sloping downward). So lets find the slope of this line:


*[invoke slope -2, 5, 4, -1]


Now lets find the slope of the line going through (-2,-1) and (4,5)


*[invoke slope -2, -1, 4, 5]


So we have one slope of {{{-1}}} and another of {{{1}}}. Since these slopes are negative reciprocals of each other (ie {{{1=-(-1/1)}}} and {{{-1=-(1/1))}}}, these two lines are perpendicular.


#32

Lets solve {{{2x -3y =5}}} first


*[invoke graphing_linear_equations "standard", 2, -3, 5, 2, 1]


So now we have the equation


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


So this means the perpendicular slope is the negative reciprocal of {{{2/3}}}. So the negative reciprocal of {{{2/3}}} is {{{-3/2}}}. So we have a line with a slope of {{{-3/2}}} and goes through (-2,6)


*[invoke find_line_by_slope_and_point -2, 6, -1.5]


So the equation is {{{y=(-3/2)x+3}}}