# Geometry: Line EquationParallelogram ABCD has peaks: A (3,5), B (5.4) and the center M (7.8). To find and write equations of the sides and diagonals of the parallelogram, how should I do?

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You may approach the problem in the following way, such that:

NP is a line that is parallel to BC and AD

N,P represent the midpoints of the lines AB and CD

You may find the midpoint N (on AB) such that:

`2x_N = x_A + x_B => 2x_N = 3 + 5 => 2x_N = 8 => x_N = 4`

`2y_N = y_A + y_B => 2y_N = 5 + 4 => 2y_N = 9 => y_N = 9/2`

You need to find the slope of the line MN, such that:

`m_(MN) = (y_M - y_N)/(x_M - x_N)`

`m_(MN) = (8 - 9/2)/(7 - 4) => m_(MN) = 7/6`

You need to remember that the slopes of two parallel lines are equal, hence, since the MN || AD yields:

`m_(MN) = m_(AD) = 7/6`

You may write the equation of the side AD, such that:

`y - y_A = m_(AD)(x - x_A) => y - 5 = 7/6(x - 3)`

`y = 5 + 7/6x - 7/2 => y = 7/6x + 3/2`

You may write the equation of the side BC, such that:

`y - y_B = m_(AD)(x - x_B) => y - 4 = 7/6(x - 5)`

`y = 7/6x - 35/6 + 4`

`y = 7/6x - 11/6`

You need to find the cordinates C and D, using the midpoint property, such that:

`x_C = 2x_M - x_A => x_C = 11 `

`y_C = 2y_M - y_A => y_C = 11`

`x_D = 2x_M - x_B => x_D = 9`

`y_D = 2y_M - y_B => y_D = 12`

You may write the equations of diagonals AC and BD, such that:

`y - y_A = (y_C - y_A)/(x_C - x_A)(x - x_A)`

`y - 5 = (11 - 5)/(11 - 3)(x - 3)`

`y - 5 = (3/4)(x - 3)`

`y - 4= (y_D - y_B)/(x_D - x_B)(x - 5)`

`y - 4= (12 - 4)/(9 - 5)(x - 5)`

`y - 4 = 2(x - 5) => y = 2x - 6`

**Hence, evaluating the equations of the sides AD, BC and diagonals AC and BC yields: **`y = 7/6x + 3/2 ; y = 7/6x - 11/6 ; y - 5 = (3/4)(x - 3) ; y = 2x - 6.`

The equation of the line AB, where A (3,5), B (5,4) :

(x-xA)/(xB-xA)=(y-yA)/(yB-yA)

(x-3)/(5-3)=(y-5)/(4-5)

(x-3)/(2)=(y-5)/(-1)

By cross multiplying:

(-1)(x-3)=2(y-5)

3-x=2y-10

2y=13-x

**yAB=(13-x)/2**

For finding the coordinates of the C point, we'll use the rule of the coordinates of the center pont M(7,8)

xM=(xA+xC)/2

7=(3+xC)/2, 14-3=xC, **xC=11**

yM=(yA+yC)/2

8=(5+yC)/2

16-5=yC

**yC=11**

For finding the coordinates of the D point, we'll use the rule of the coordinates of the center pont M(7,8)

xM=(xB+xD)/2

7=(5+xD)/2

14-5=xD

**xD=9**

yM=(yB+yD)/2

8=(4+yD)/2

**yD=12**

The equation of the line BC, where C (11,11), B (5,4) :

(x-xB)/(xC-xB)=(y-yB)/(yC-yB)

(x-5)/(6)=(y-4)/(7)

7(x-5)=6(y-4)

7X-35=6Y-24

7x-11=6y

**yBC=(7x-11)/6**

The same you will calculate for the lines AD and CD.

The diagonal of the parallelogram are AC and BD.

Equation of AD: The diagonal line pass through A (3,5) and the point D not given must pass through the centre M (7,8)

Therefore, the equation of the line is: (x-3)/(3-7)=(y-5)/(5-8) or (x-3)(-3)=(y-5)(-4) or 3x-9=4y-20, which simplifies to 3x-4y+11=0.

The equation of the other diagonal BD:

BD passes through B (5,4) and M(7,8). So its equation is:

(x-5)/(5-7)=(y-4)/(4-8) or (x-5)(-4)=(y-4)(-3) or 4x-20=3y-12, which simplifies to 4x-3y-8=0

Since M is the mid point of AC and BD,the coordinates of C (x,y) could be found by equating the algebraic mid point of AC with that of M.

((3+x)/2 , (5+y)/2 ) = coordinates of M (7,8). We get C (11,11).Similarly try and find D by treatind M is mid point of B and D. So, we find D(9,12).

The equation of AD is the line joining A(3,5) and (9,12).

Which is y-12= [12-5)/(9-3)]{(x-9)

y-12=(7/6)(x-9) or

7x-6y+9=0 is the equation of AD.

Equation of BC , which joins B (5,4) and C( 11,11) is

y-11 = ((11-4)/(11-5))(x-11) =(7/6)(x-11) or

7x-6y-11 = 0.

Similarly you can find equations for the nes AB and CD which are the other two parallel sides of the parallelogram ABCD.