Question 1141667
 
given:
The vertices of a triangle ABC 

A({{{2}}},{{{4}}}) 
B({{{-2}}},{{{0}}}) 
 C({{{6}}},{{{-1}}})


(a) Find the coordinate of the midpoint P, Q, R of the sides AB, BC and CA respectively.

The midpoint of two points, ({{{x[1]}}}, {{{y[1]}}}) and ({{{x[2]}}}, {{{y[2]}}}) is the point M found by the following formula:

M= ({{{(x[1] + x[2])/2}}},{{{(y[1] + y[2])/2}}})
​
the coordinates of the midpoint P, of the sides AB are:
 
P = ({{{(2+ (-2))/2}}},{{{(4 + 0)/2}}})
​P = ({{{0/2}}},{{{4 /2}}})
	
P = ({{{0}}},{{{2}}})


the coordinates of the midpoint Q, of the sides BC are:

Q=({{{(-2 + 6)/2}}},{{{(0 -1)/2}}})
Q=({{{2}}},{{{-1/2}}})

the coordinates of the midpoint R, of the sides CA are:

R= ({{{(2+ 6)/2}}},{{{(4-1)/2}}})
R= ({{{4}}},{{{3/2}}})



(b) Find the equation of the medians AQ and BR, and the coordinates of the point were they intersect.
the equation of the medians:

AQ =>

first use points A({{{2}}},{{{4}}})  and Q=({{{2}}},{{{-1/2}}}) to find a slope

{{{m=(y[2]-y[1])/(x[2]-x[1])}}}

{{{m=(-1/2-4)/(2-2)}}}

{{{m=(-1/2-4)/0}}}=> slope is undefined, and an undefined slope indicates that we have a vertical line parallel to the y-axis and passing through all points in the plane with an x-coordinate = constant ( c)

since both points {{{A}}} and {{{Q}}} have same x-coordinate, the equation of the median is

{{{x=2}}}


 
and BR

B({{{-2}}},{{{0}}}) 

R= ({{{4}}},{{{3/2}}})

{{{m=(3/2-0)/(4-(-2))}}}

{{{m=(3/2)/(4+2)}}}

{{{m=(3/2)/6}}}....simplify

{{{m=(cross(3)1/2)/cross(6)2}}}

{{{m=1/4}}}


now use point slope formula: 


{{{y-y[1]=m(x-x[1])}}}......plug in a slope and the coordinates of one point

{{{y-0=(1/4)(x-(-2))}}}

{{{y=(1/4)(x+2)}}}

{{{y=(1/4)x+2(1/4)}}}

{{{y=(1/4)x+1/2}}}



now find the coordinates of the point were they intersect:

{{{x=2}}}........eq.1
{{{y=(1/4)x+1/2}}}.......eq.2
------------------------------------

{{{y=(1/4)2+1/2}}}

{{{y=1/2+1/2}}}

{{{y=1}}}


so, the coordinates of the point were they intersect is  I=({{{2}}},{{{1}}})

(c) Show that the third median CP also passes through this point 

 the third median CP:

 C({{{6}}},{{{-1}}})
P = ({{{0}}},{{{2}}})

first find the equation of the median:

{{{m=(2-(-1))/(0-6)}}}

{{{m=(2+1)/(-6)}}}

{{{m=3/(-6)}}}

{{{m=-1/2}}}

{{{y-y[1]=m(x-x[1])}}}......plug in a slope and the coordinates of one point

{{{y-2=-(1/2)(x-0)}}}

{{{y-2=-(1/2)x}}}

{{{y=-(1/2)x+2}}}


plug in coordinates I=({{{2}}},{{{1}}})

{{{1=-(1/2)2+2}}}

{{{1=-1+2}}}
{{{1=1}}}=> which is true and confirms that the third median CP also passes through the point I=({{{2}}},{{{1}}})



{{{drawing( 600, 600, -10, 10, -10, 10,

line(2,10,2,-10), circle(2,4,.12),circle(-2,0,.12),circle(6,-1,.12),
locate(2,4,A), locate(-2,-0.5,B),locate(6,-1,C),circle(0,2,P),locate(0,2,P),
circle(2,-1/2,Q),locate(2,-1/2,Q),
circle(4,3/2,R),locate(4,3/2,R),circle(2,1,.12),locate(2,1,I),
green(line(2,4,-2,-0)), green(line(2,4,6,-1)), green(line(-2,0,6,-1)),
 graph( 600, 600, -10, 10, -10, 10, -(1/2)x+2,(1/4)x+1/2)) }}}