Question 98441
1)


"Find the length  of AB"

Start with the given distance formula

{{{d=sqrt((x[1]-x[2])^2+(y[1]-y[2])^2)}}} where *[Tex \Large \left(x_{1}, y_{1}\right)] is the first point *[Tex \Large \left(-10,6\right)] and *[Tex \Large \left(x_{2}, y_{2}\right)] is the second point *[Tex \Large \left(4,2\right)]


{{{d=sqrt((-10-4)^2+(6-2)^2)}}} Plug in {{{x[1]=-10}}}, {{{x[2]=4}}}, {{{y[1]=6}}}, {{{y[2]=2}}}


{{{d=sqrt((-14)^2+(4)^2)}}} Evaluate {{{-10-4}}} to get -14. Evaluate {{{6-2}}} to get 4. 


{{{d=sqrt(196+16)}}} Square each value


{{{d=sqrt(212)}}} Add


{{{d=2*sqrt(53)}}} Simplify the square root  (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)



So the distance approximates to


{{{d=14.560219778561}}}


which rounds to

14.56


So the distance between (-10,6) and (4,2) is approximately 14.56 units



-----------------------------------------------------------------



"Find the midpoint of ab"



In order to find the midpoint, we need to average each corresponding coordinate. In other words, we need to add up the corresponding coordinates and divide the sum by 2.



So lets find the averages between the two points




To find *[Tex \Large  \textrm{x_{mid}}], average the x-coordinates between the two points

{{{x[mid]=(-10+4)/2=(-6)/2=-3}}}



So the x-coordinate of the midpoint is -3 (i.e. x=-3)

-----------------------------------------------------------------------------------------------------------------



To find *[Tex \Large  \textrm{y_{mid}}], average the y-coordinates between the two points

{{{y[mid]=(6+2)/2=(8)/2=4}}}



So the y-coordinate of the midpoint is 4 (i.e. y=4)

-----------------------------------------------------------------------------------------------------------------

Answer:

Since the coordinates of the midpoint are x=-3, y=4, this means the midpoint is (-3,4)


Check:

Here is a graph to visually see the answer

{{{drawing(500, 500, -12, 6, 0, 8,
graph(500, 500, -12, 6, 0, 8,0),
blue(line(-10,6,4,2)),
circle(-10,6,0.05),
circle(-10,6,0.08),
circle(4,2,0.05),
circle(4,2,0.08),
circle(-3,4,0.05),
circle(-3,4,0.08)

)}}} Graph of the line segment with the endpoints (-10,6) and (4,2) with the midpoint (-3,4)




We could visually verify our answer if we simply draw right triangles from each point like this:

{{{drawing(500, 500, -12, 6, 0, 8,
graph(500, 500, -12, 6, 0, 8,0),
blue(line(-10,6,4,2)),

line(-10,6,-10,(6+2)/2),
line(-10,(6+2)/2,-3,4),
line(-3,4,(-10+4)/2,2),
line((-10+4)/2,2,4,2),

circle(-10,6,0.05),
circle(-10,6,0.08),
circle(4,2,0.05),
circle(4,2,0.08),
circle(-3,4,0.05),
circle(-3,4,0.08)

)}}}


Here we can see that the two triangles are congruent, so our answer is verified.




<hr>



2)


"Find the length  of AB"



Start with the given distance formula

{{{d=sqrt((x[1]-x[2])^2+(y[1]-y[2])^2)}}} where *[Tex \Large \left(x_{1}, y_{1}\right)] is the first point *[Tex \Large \left(8,3\right)] and *[Tex \Large \left(x_{2}, y_{2}\right)] is the second point *[Tex \Large \left(7,-2\right)]


{{{d=sqrt((8-7)^2+(3--2)^2)}}} Plug in {{{x[1]=8}}}, {{{x[2]=7}}}, {{{y[1]=3}}}, {{{y[2]=-2}}}


{{{d=sqrt((1)^2+(5)^2)}}} Evaluate {{{8-7}}} to get 1. Evaluate {{{3--2}}} to get 5. 


{{{d=sqrt(1+25)}}} Square each value


{{{d=sqrt(26)}}} Add


So the distance approximates to


{{{d=5.09901951359278}}}


which rounds to

5.1


So the distance between (8,3) and (7,-2) is approximately 5.1 units



------------------------------------------------------------------------------



"Find the midpoint of ab"



In order to find the midpoint, we need to average each corresponding coordinate. In other words, we need to add up the corresponding coordinates and divide the sum by 2.



So lets find the averages between the two points




To find *[Tex \Large  \textrm{x_{mid}}], average the x-coordinates between the two points

{{{x[mid]=(8+7)/2=(15)/2=7.5}}}



So the x-coordinate of the midpoint is 7.5 (i.e. x=7.5)

-----------------------------------------------------------------------------------------------------------------



To find *[Tex \Large  \textrm{y_{mid}}], average the y-coordinates between the two points

{{{y[mid]=(3+-2)/2=(1)/2=0.5}}}



So the y-coordinate of the midpoint is 0.5 (i.e. y=0.5)

-----------------------------------------------------------------------------------------------------------------

Answer:

Since the coordinates of the midpoint are x=7.5, y=0.5, this means the midpoint is (7.5,0.5)


Check:

Here is a graph to visually see the answer

{{{drawing(500, 500, 5, 10, -4, 5,
graph(500, 500, 5, 10, -4, 5,0),
blue(line(8,3,7,-2)),
circle(8,3,0.05),
circle(8,3,0.08),
circle(7,-2,0.05),
circle(7,-2,0.08),
circle(7.5,0.5,0.05),
circle(7.5,0.5,0.08)

)}}} Graph of the line segment with the endpoints (8,3) and (7,-2) with the midpoint (7.5,0.5)




We could visually verify our answer if we simply draw right triangles from each point like this:

{{{drawing(500, 500, 5, 10, -4, 5,
graph(500, 500, 5, 10, -4, 5,0),
blue(line(8,3,7,-2)),

line(8,3,8,(3+-2)/2),
line(8,(3+-2)/2,7.5,0.5),
line(7.5,0.5,(8+7)/2,-2),
line((8+7)/2,-2,7,-2),

circle(8,3,0.05),
circle(8,3,0.08),
circle(7,-2,0.05),
circle(7,-2,0.08),
circle(7.5,0.5,0.05),
circle(7.5,0.5,0.08)

)}}}


Here we can see that the two triangles are congruent, so our answer is verified.