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Question 1185154: 1.The points A(-7, -7), B(8, -1), C(4, 9), D are the vertices of the parallel ogram ABCD. Find the coordinates of D. Prove that ABCD is a rectangle and find its area.
b. Find the equation of the line which is parallel to the line 3x + 4y = 12 and which makes an intercept of 5 units on the x-axis. Find also the equation of the line which is perpendicular to the given line and which passes through the point (4, 5).
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 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)
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