can someone help me with coordinate problems?

Apr 2010
22
0
Given: A(1, -8), B(5, 4). M is the midpoint of AB

A) describe fully the locus of points equidistant from A and B
B) find the slope of AB
C) find the coordinate of M
D) write an equation of the locus of points equidistant from A and B
E) find the length of AM
F) if AB is the diameter of a circle write the equation of that circle
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
Given: A(1, -8), B(5, 4). M is the midpoint of AB

A) describe fully the locus of points equidistant from A and B

perpendicular bisector of AB

B) find the slope of AB

look up the formula for the slope between two points

C) find the coordinate of M

look up the midpoint formula

D) write an equation of the locus of points equidistant from A and B

use the coordinates of point M and the slope perpendicular to AB

E) find the length of AM

look up the distance formula

F) if AB is the diameter of a circle write the equation of that circle

standard equation for a circle ...

\(\displaystyle \textcolor{red}{(x - h)^2 + (y-k)^2 = r^2}\)

... where (h,k) is the circle's center
...
 
Apr 2010
22
0
can someone check if i answer the question correctly?

for slope i did

4 - -(8) / 5 - 1 = 12 / 4 = 3

so slope is 3

for midpoint i got

1 + 5 / 2 = 2
-(8) + 4 / 2 = -(2)

so coordinate is (2,-2)

for the distance i got

A (1,-8) M(2,-2)

2 - 1 square + -(2) - 8
1 + 100 = 101 and square root i got 10.04

the equation of locus i got

y = 3x + -1

and the circle equation i am confused what is h and k?
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
can someone check if i answer the question correctly?

for slope i did

4 - -(8) / 5 - 1 = 12 / 4 = 3 correct

so slope is 3

for midpoint i got

(1 + 5) / 2 = 2 ??? ... recheck this x calculation

-(8) + 4 / 2 = -(2)

because the x-value of the midpoint is incorrect, you'll need to recompute the remainder of these questions.

and the circle equation i am confused what is h and k?

(h,k) is the circle's center (the midpoint)
...
 
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Apr 2010
22
0
oh no wonder it looks wrong thx for checking

for midpoint i got (3 , -2)

distance i got 40 square root it i can do 2 squaure root 20

the equation i got y = -3x + 1

and the circle equation i got ( x - 9 ) + ( y - -(4) ) = is r = 40?
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
oh no wonder it looks wrong thx for checking

for midpoint i got (3 , -2) ok

distance i got 40 square root it i can do 2 squaure root 20

\(\displaystyle \textcolor{red}{\sqrt{40} = 2\sqrt{10}}\)

the equation i got y = -3x + 1

no ... the slope of a perpendicular is the opposite reciprocal.

and the circle equation i got ( x - 9 ) + ( y - -(4) ) = is r = 40?

\(\displaystyle \textcolor{red}{r^2 = 40}\)

\(\displaystyle \textcolor{red}{(x-3)^2 + (y+2)^2 = 40}\)
...
 
Apr 2010
22
0
oh i changed to reprocial and for the slope i got y = 1/3 x + -1
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
oh i changed to reprocial and for the slope i got y = 1/3 x + -1
perpendicular slope is \(\displaystyle m = -\frac{1}{3}\)

\(\displaystyle y - (-2) = -\frac{1}{3}(x - 3)
\)

\(\displaystyle y + 2 = -\frac{1}{3}x + 1\)

\(\displaystyle y = -\frac{1}{3}x - 1\)
 
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