Question

The point (1,2) and (3,8) are pair of opposite vertices of a square.find the equation of the sides and diagonal of a square and diagonal of the square

Answer 1

Sides equation : y = -2x + 4 , 2y  = x + 3 , y = -2x + 14 , 2y  = x + 13

Diagonal equation :  y = 3x - 1   &   3y + x = 17

Step-by-step explanation:

Let say one of the remaining vertex  = (x , y)

Diagonal = Side√2

=> Side² = Diagonal²/2

then

(x - 1)² + (y - 2)² = (x - 3)²  + (y - 8)²   = ((8 - 2)² + (3 - 1)² )/2

=> x²  + y²  - 2x - 4y + 5 = x² + y² -6x - 16y + 73  =  20

x²  + y²  - 2x - 4y + 5 = x² + y² -6x - 16y + 73

=> 4x + 12y = 68

=> x + 3y  = 17

x = 17 - 3y

x²  + y²  - 2x - 4y + 5 =  20

=> (17 - 3y)² + y²  - 2(17 - 3y) - 4y + 5 =  20

=> 289 + 9y² - 102y  + y² - 34 +6y - 4y + 5 = 20

=> 10y² - 100y + 240 =0

=> y² - 10y + 24 =0

=> y² - 4y - 6y + 24 = 0

=> (y - 4)(y - 6) =0

=> y  = 4   or 6

x = 5   or  - 1

Points of square are

(- 1 , 6)  ( 1 , 2)  , (5 , 4)  , ( 3, 8)

Equation of sides (- 1 , 6)  ( 1 , 2)

Slope = (-4/2) = - 2  => y = -2x + c

2 = - 2*1 + c => c = 4

=> y = -2x + 4

Equation of sides  ( 1 , 2)  , (5 , 4)

Slope = (2/4) = 1/2 => y = x/2  + c

2  = 1/2 + c => c = 3/2

=> y = x/2 + 3/2  => 2y  = x + 3

Equation of sides  (5 , 4) , ( 3, 8)

slope = - 2   => y = -2x + c

=> 8 = -2*3 + c => c = 14

y = -2x + 14

Equation of sides ( 3, 8) , (- 1 , 6)

slope = 1/2   => y = x/2  + c

8 = 3/2 + c  => c = 13/2

=> y = x/2 + 13/2  => 2y  = x + 13

Equation of Diagonal   (1,2) and (3,8)

slope = 6/2 = 3  => y = 3x + c

=> 2 = 3 + c => c = -1

=> y = 3x - 1

Equation of Diagonal   (- 1 , 6) , (5 , 4)

slope = -2/6 = -1/3  => y = -x/3 + c

=> 6 = 1/3 + c => c = 17/3

=>  y = -x/3  + 17/3

=> 3y + x = 17

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