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August 30th, 2018, 12:35 PM   #1
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Coordinate geometry

A square ABCD has its diagonal AC and BD lying along the lines
3x-7y-25=0
7x+3y-10=0 respectively
The side CD lies along the line
2x+5y-7=0
1. Calculate the coordinates of the triangle ACD
2. Obtain the equation of the line AB
3. Find the perimeter of ACD

I have tried substituting equ. of line1 from 2, but that does not work. I need all possible aid.... Please.

Last edited by skipjack; August 30th, 2018 at 02:13 PM.
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August 30th, 2018, 01:04 PM   #2
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1. and 3. give you point C, 2. and 3. give you point D. This gives you one side. Now you can graph it. See if you can do the rest.
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August 31st, 2018, 05:37 AM   #3
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As CD has slope -2/5, AD has slope -1/(-2/5) = 5/2.
This may assist in finding the coordinates of A.
It's then easy to find the equation of AB.

Solving 3x - 7y - 25 = 0 and 2x + 5y - 7 = 0, as suggested by mathman, gives (x, y) = (6, -1). This is point C.

Solving 7x + 3y - 10 = 0 and 2x + 5y - 7 = 0 gives x = y = 1, so D is the point (1, 1).

Solving 3x - 7y - 25 = 0 and 7x + 3y - 10 = 0 gives (x, y) = (5/2, -5/2) as the coordinates of where the diagonals of the square intersect. You don't need to know that point, but as it's the mid-point of both AC and BD, it's now very easy to find the coordinates of A and B.

Specifically, A = 2(5/2, -5/2) - (6, -1) = (-1, -4), and B = 2(5/2, -5/2) - (1, 1) = (4, -6).

You now have the coordinates of every point referred to, so you have the coordinates of at least two points on each of the lines referred to.

This article gives two ways of writing an equation of the line through two known points.
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August 31st, 2018, 06:29 AM   #4
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Quote:
Originally Posted by Harmeed View Post
A square ABCD has its diagonal AC and BD lying along the lines
3x-7y-25=0
7x+3y-10=0 respectively
The side CD lies along the line
2x+5y-7=0
1. Calculate the coordinates of the triangle ACD
2. Obtain the equation of the line AB
3. Find the perimeter of ACD

I have tried substituting equ. of line1 from 2, but that does not work. I need all possible aid.... Please.
I have nothing to add substantively to what has already been said, but if you had drawn a diagram, you would have physically seen that CD and AC intersect at C. Therefore, the equations for those two lines are equal at C. Diagrams are useful.
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September 1st, 2018, 09:15 PM   #5
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Help me, why did he multiply the coordinates of the midpoint by 2 before subtraction? That's not clear.

Last edited by skipjack; September 2nd, 2018 at 06:21 AM.
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September 2nd, 2018, 06:21 AM   #6
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For any two points R and S, their mid-point M = (R + S)/2, so 2M = R + S.

Hence subtracting either one of the original points from 2M gives the other one.
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