Question 1185154
Sketch this and the three points are in quadrants III, IV, and I. So the other point is in quadrant 2.
The slope between A and B is change in y/change in x=6/15 or 2/5.
the slope between B and C is 10/-4 or -5/2, which is the negative reciprocal of 2/5, so the lines are perpendicular. ANSWER
Now look at distances. A and B, the long side, distance is sqrt (15^2+6^2)=sqrt(261).
Need the same distance from C to D. 
Call D (x, y)
Then sqrt(( x-4)^2+(9-y)^2)=sqrt(261).
the x distance has to be 15, so it is x=-11.
The y distance has to be 6, so it is y=3. Only that way will the distance be sqrt(261) Again, check the sketch.
D is (-11, 3); A is (-7, -7) ANSWER
Prove this is a rectangle by looking at the slope between A and D. It should be -5/2. It is =-10/4, so that is OK.
Now look at the distance between A and D, and that is sqrt (16+100)=sqrt (116).
Compare with the parallel lie BC, which has distance sqrt(16+100)=sqrt (116)
The lines are parallel to each other, perpendicular to each other, and opposite sides are of equal length. That is a rectangle.
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The area, in square units, is sqrt(261)*sqrt(116)=174 square units. ANSWER
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3x+4y=12
4y=-3x+12
y=-(3/4)x+3
parallel line has slope of -3/4 and point at (5, 0)
point slope formula, y-y1=m(x-x1), m slope (x1, y1) a point,
y=-(3/4)(x-5)
y=-(3/4)x+(15/4) ANSWER
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perpendicular line to the given line has slope of (4/3) and goes through (4, 5)
y-5=(4/3)(x-4)
y=(4/3)x-(1/3)
{{{graph(300,300,-5,5,-5,5,(-3/4)x+3,-(3/4)x+3.75,(4/3)x-(1/3))}}}