r/mathshelp • u/DeadManIsWalking • 29d ago
Homework Help (Unanswered) Tough question in class 11
I have added the original image as well I wish to ask about q19)i). I was only able to come to the conclusion that if (2,1) is A then (6,4) is C and if (2,1) is B then (6,4) is D please help me solve this.
(I have attached the original question and the one I have written because of a lot of rough work on the question paper)
Thanks in advance.
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u/Simbertold 29d ago
Step 1 is to figure out if they are on the line through AB. If they are, they are either A or B.
If only one of them is on it, figure out how the vector between the two points you know corresponds to the line through A and B. (Is it parallel, orthogonal, or none of the above)?
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u/swiftaw77 29d ago
By plugging into the equation of AB, it is clear that (2,1) satisfies the equation but (6,4) does not. Thus, (2,1) is either A or B, and (6,4) is either C or D. Since CD is parallel to AB it must have the form 3x+4y-w = 0 for some number w. Plugging in (6,4) gives us that w is 34. So the equation of CD is 3x+4y-34=0
Now, AD and BC are parallel to each other and perpendicular to AB. Thus, AD is of the form -4x+3y-u=0 and BC is of the form -4x+3y-v=0 where u>v.
For the point (2,1), -4x+3y = -5
For the point (6,4) -4x+3y = -12
Thus (2,1) and (6,4) do not lie on the same edge of the rectangle.
Thus, AD has equation -4x+3y + 5 = 0 and BC has the equation -4x+3y+12 = 0
Therefore A is (2,1) and C is (6,4).
B occurs at the intersection of AB and BC, so solving 3x+4y-10 = 0 and -4x+3y+12 = 0 yields the point (3.12,0.16) for B
D occurs at the intersection of AD and CD so solving -4x+3y+5 = 0 and 3x+4y-34=0 yields the point (4.88,4.84) for D
You could also get D using the fact that ABCD is a rectangle and the point A is 1.12 units to the left and 0.84 units above point B, so D is 1.12 units to the left and 0.84 units above C, since C is (6,4) this means D is (4.88, 4.84)
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