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
Answers
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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