Question

A(1,7) B(6,4) and C(5,5) are three points in a plane

Answer 1

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 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)