Question 1126749
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Please help me solve this:


"AB has endpoints A(-5,0) and B(4,3). CD has endpoints C(-3,9) and D(1,-3). The equations of the lines 
containing AB and CD are x - 3y = -5 and 3x + y = 0, respectively."



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a. How could you quickly check that these equations are correct?


   The simplest way to check that AB lies on the straight line  x - 3y = -5  is to check that each endpoint A and B lies on  this line.


    To check it for AB, simply substitute the coordinates of  A  x= -5  and y= 0 into the equation:

    -5 -3*0 = - 5  (! correct !),   


    and  then substitute the coordinates of  B  x= 4  and y= 3  into the equation:

    4 -3*3 = - 5  (! correct !).


    Thus AB is checked.   Now you check  for CD  by the same way.



b. Verify that the lines are perpendicular.


   The line x - 3y = -5 has the slope  {{{1/3}}}.

   The line  3x + y = 0  has the slope  -3.


   The numbers  {{{1/3}}} and -3 are opposite reciprocal.

   It means that the two lines are perpendicular.



c. Find the point of intersection of AB and CD by solving the system of equations.


    x  - 3y = -5.    (1)
    3x +  y =  0.    (2)


    From eq(1) express x = -5 + 3y and substitute it into eq(2). You will get

        3*(-5+3y) + y = 0,

        -15 + 9y + y = 0

         10y = 15  ====>  y = 15/10 = 1.5.

         Then  x = -5 + 3y = -5 +3*1.5 = -5 + 4.5 = -0.5.


    Thus the intersection point is  (x,y) = (-0.5,1.5)



d. Find the midpoints of AB and CD. Compare your results with Part c.


   To find midpoint of AB,  take x-coordinates of A and B, add them and divide the sum by 2 
   (so you will get the average of x-coordinates of endpoints A and B).

   By doing it, you will get x-coordinate of the midpoint AB as  {{{(-5+4)/2}}} = -0.5.


    Now make the same with y-coordinates of A and B:  you will get y-coordinate of the midpoint AB as  {{{(0+3)/2}}} = 1.5.


    Thus the midpoint of AB  is  (x,y) = (-0.5,1.5).


    To determine the midpoint of CD, do the same with the points  C  and  D.

     Notice that is is <U>THE SAME point as the intersection point of AB and CD</U>, which we determined in  Part c). 



e. What kind of quadrilateral is ACBD? Explain your reasoning.

   The quadrilateral ABCD has perpendicular diagonals  AB  and  CD;  and the diagonals intersetion is the midpoint of each diagonal.

    Hence the quadrilateral ABCD  is a rhombus.
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COMPLETED.