Question

Let A(-7,4) and B(5,-12) be points on a plane.
Find the:
a) length of segment
b) equation of the perpendicular bisector of AB
c) equation of the circle for which AB is a diameter

Answer 1

Length of segment= sqrt{(5+7)^2 + (-12-4)^2}
=sqrt{144+256)
=20 unit

Answer 2

(a) Length is given by the Pythagorean theorem:\[\sqrt{(5-(-7))^2+(-12-4)^2}\](b) Slope is the inverse reciprocal of the slope of the line connecting the two points and will intersect the midpoint between the two points. Thus, by point-slope form, which is left for you to simplify:\[y-\left(\dfrac{4+(-12)}{2}\right)=-\dfrac{5-(-7)}{-12-4}\left[x-\left(\dfrac{-7+5}{2}\right)\right]\](c) Given by the transformation of the generic formula for a circle given that the midpoint of the two given points is the center of the circle and the radius is half the length of the line connecting them, left for you to simplify:\[\left[\dfrac{\sqrt{(5-(-7))^2+(-12-4)^2}}{2}\right]^2=\left[y-\left(\dfrac{4+(-12)}{2}\right)\right]^2+\left[x-\left(\dfrac{-7+5}{2}\right)\right]^2\]

Answer 3

eqn of perpendicular bisector ---
first slope of line AB =- (-12-4)/(5+7)
=4/3
then slope perpendicular bisector of AB=-3/4
and mid pt of line AB={(-7+5)/2 , (4-12)/2}
=(-1,-4)

so eqn of perpendicular bisector of AB will be
y+4=-3/4(x+1) {using y-y1=m(x-x1)}
4y+4=-3x-3
=> 3x+4y + 7=0

Answer 4

it says teh asnwer is 3x-4y = 13

Answer 5

for teh equation of the perpemdicular bisector i mean

Answer 6

did you get the length of the line segment?

Answer 7

yeah i understand taht...its 20 with the distance formula. i'm stuck on the otehr 2 now

Answer 8

\[\sqrt{(5-(-7))^2+(-12-4)^2}=\sqrt{12^2+16^2}=20\]

Answer 9

thanks taht shows all teh steps

Answer 10

you need the midpoint right?

Answer 11

yeah

Answer 12

\[(\frac{-7+5}{2},\frac{4-12}{2})\]
\[(-1,-4)\]

Answer 13

is that ok? average of the coordinates

Answer 14

yeah thanks

Answer 15

ok now the equation for the circle. center is
\[(-1,-4)\] radius is 10 (half of 20) so equation for the circle is
\[(x+1)^2+(y+4)^2=10^2\]

Answer 16

or if you prefer
\[(x+1)^2+(y+4)^2=100\]

Answer 17

now for the equation of the perpendicular bisector

Answer 18

(à la what I wrote above)

Answer 19

first we need the slope of the line between the two points
\[(-7,4) , (5,-12)\]
it is
\[m=\frac{-12-4}{5+7}=\frac{-16}{12}=-\frac{4}{3}\]

Answer 20

so the slope of the perpendicular line is
\[\frac{3}{4}\] clear so far?

Answer 21

yeah its negative recipricol

Answer 22

ok and so we have the slope, and also the point, namely
\[(-1,-4)\] so use the all mighty point slope formula
\[y-y_1=m(x-x_1)\]so get the equation. it is
\[y+4=\frac{3}{4}(x+1)\]

Answer 23

now if you want to write this in some other form, say slope - intercept for or standard form you can change it however you like. for example you could write
\[y+4=\frac{3}{4}x+\frac{3}{4}\]
\[y=\frac{3}{4}x-\frac{13}{4}\]

Answer 24

how did they get teh answer 3x-4y = 13? what format is that

Answer 25

dumb math teacher form, but we can get there too

Answer 26

lol..yupp math teachers liek to make everything look and sound complicated

Answer 27

\[y=\frac{3}{4}x-\frac{13}{4}\]
\[4y=3x-13\]
\[13=3x-4y\] whatever you like

Answer 28

\[3x-4y=13\]
\[3x-4y-13=0\] anything

Answer 29

wasn't that bad was it?

Answer 30

no it actually wasnt. =) except when see teh question at first it drives me crazy

Answer 31

i have a question. For the equation 9x^2 + 16y^2 = 144 is it posible to make it into
y=mx+b form

Answer 32

heck no

Answer 33

so how wud u draw a graph for that

Answer 34

for one thing
\[y=mx+b\] is a line and you have an ellipse!
http://www.wolframalpha.com/input/?i=9x^2+%2B+16y^2+%3D+144

Answer 35

oh i see...so how did tehy get those points or liek how did they draw the ellipse

Answer 36

like for a circle u have a center n a radius so u can draw it...how about this? what do each of teh numbers mean

Answer 37

there is no snap answer to this question. you can write it in the form
\[\frac{x^2}{16}+\frac{y^2}{9}=1\] and then look at the standard form on the ellipse.
http://mysite.du.edu/~jcalvert/math/ellipse.htm

Answer 38

do u sqt root the values of 9 and 16 b.c in teh drawing on teh x values it reaches close to 4 and on teh y value it is close to 3

Answer 39

im confused how u sketch a ellipse with teh equation 9x^2 + 16y^2 = 144

Answer 40

@august151985....r u there