Answer:
Step-by-step explanation:
Middle point of AB
x(m) = (6+1)/2 = 7/2
y(m) = (7+4)/2 = 11/2
slope of the line that contains AB
(4-7)/(6-1) = -3/5
eqaution of the perpendicular bisector
y-11/2 = 5/3(x-7/2)
y = 5/3x -35/6 + 11/2
y = 5/3x + (-35 + 33)/6
y = 5/3x -1/3
Middle point of AC
x(m) = (1+5)/2 = 3
y(m) = (7+5)/2 = 6
Slope of the line that contains AC
(5-7)/(5-1) = -1/2
equation of the perpendicular bisector
y-6 = 2(x-3)
y = 2x -6 + 6
y = 2x
Point of intersection
y= 5/3x -1/3
y = 2x
2x = 5/3x - 1/3
6x = 5x - 1
x = -1
y = -2
P(-1,-2)
Answer:
(1,7)
Step-by-step explanation:
Given:
A(1,7)
B(6,4)
C(5,5)
Solution:
Mid point of AB = M((1+6)/2,(7+4)/2) = M(3.5,5.5)
Slope of AB = (4-7)/(6-1) = -3/5
Perpendicular bisector of AB:
L1: y - 11/2 = -(3/5)(x-7/2) ............(1)
Mid point of AC, m= N((1+5)/2,(7+5)/2) = N(3,6)
Slope of AC, n = (5-7)/(5-1) = -2/4 = -1/2
perpendicular bisector of AC:
L2: y-6 = -(1/2)(x-3) ..........."(2)
To find the point of intersection,
(1)-(2)
-5.5 - (-6) = -(3/5)x +12/5 + x/2 - 3/2
1/2 = -x/10 + 6/10
x/10 = 1/10
x = 1
substitute x in (1)
y = 3/2+11/2 =7
Therefore Point of intersection is (1,7)
