the endpoints of AB¯¯¯¯¯ are A(-3, 4) and B(1, 2). Line k is the perpendicular bisector of AB¯¯¯¯¯. Determine the equation of line k in slope-intercept form (write the equation without spacing).
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Answers
Given : the endpoints of AB are A(-3, 4) and B(1, 2).
Line k is the perpendicular bisector of AB ¯.
To Find : The equation of line k in slope-intercept form
Solution:
Method 1 : any point on perpendicular bisector of AB will be Equidistant from A and B
so find locus of P (x , y) such that AP = BP
=> AP² = BP²
Using distance formula
=> (x - (-3))² + (y - 4)² = (x - 1)² + (y - 2)²
=> x² + 6x + 9 + y² - 8y + 16 = x² -2x + 1 + y² - 4y + 4
=> 8x -4y = -20
=> 4y = 8x + 20
=> y = 2x + 5
Method 2 :
Slope of AB = (2 - 4)/(1 - (-3))
= -2/4
= -1/2
Slop of perpendicular line = -1/(-1/2) = 2
Mid point of AB = (-3 + 1)/2 , ( 4 + 2)/2 = -1 , 3
y - 3 = 2 (x - (-1))
=> y -3 = 2x + 2
=> y = 2x + 5
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Answer:
Method 1: any point on perpendicular bisector of AB will be Equidistant from A and B
so find locus of P (x, y) such that AP = BP
=> AP² = BP²
Using distance formula
=> (x - (-3))² + (y - 4)² = (x - 1)² + (y - 2)²
=> x² + 6x + 9 + y² - 8y + 16
= x² -2x + 1 + y² -4y + 4
=> 8x -4y = -20
=> 4y = 8x + 20
=> y = 2x + 5
Method 2:
Slope of AB = (2 - 4) / (1 - (- 3))
= -2/4
= -1/2
Slop of perpendicular line = -1 :-1/(-1/2) = 2
Mid point of AB = (-3+1)/2, (4 + 2)/2 = -1,3
y - 3 = 2(x - (- 1))
=> y -3 = 2x + 2
=> y = 2x + 5.