circle O and circle P intersect at A and B. Show that OP is the perpendicular bisector of the common chord AB

Step-by-step explanation:

Let two circles O and O' intersect at two points A and B so that AB is the common chord of two circles and OO' is the line segment joining the centres

Let OO

′

intersect AB at M

Now Draw line segments OA,OB,O

′

AandO

′

B

In ΔOAO

′

andOBO

′

 , we have

OA=OB (radii of same circle)

O

′

A=O

′

B (radii of same circle)

O

′

O=OO

′

 (common side)

⇒ΔOAO

′

≅ΔOBO

′

(SSS congruency)

⇒∠AOO

′

=∠BOO

′

⇒∠AOM=∠BOM(i)

Now in ΔAOM and ΔBOM we have

OA=OB (radii of same circle)

∠AOM=∠BOM (from (i))

OM=OM (common side)

⇒ΔAOM≅ΔBOM (SAS congruncy)

⇒AM=BM and∠AMO=∠BMO

But

∠AMO+∠BMO=180°

⇒2∠AMO=180°

⇒∠AMO=90°

Thus,AM=BM and∠AMO=∠BMO=90°

HenceOO is the perpendicular bisector of AB.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

this may not be at all, but i use the same website you're using for school, and if i'm stuck doing one of these i try to look for it on the section.