Question 683785
(6,0), (5,3), and (-1,-4)

Let A (6,0), B(5,3), and C(-1,-4)be the vertices of the triangle

Let P,Q be the mid points of AB & AC respectively

PQ is parallel to BC

Determine co ordinates of P & Q by mid point formula

A (6,0), B(5,3), 

Let P(x3,y3)

{{{x3= (6+5)/2 = 11/2}}}
{{{y3=(3+0)/2 = 3/2}}}

p=((11/2),(3/2))

Similarly find co ordinates of Q the mid point of AC

Q=(x4,y4)
A (6,0), C(-1,-4)
{{{x4= (-1+6)/2=(5/2)}}}
{{{y4=(-4+0)/2= -2}}}

Q=((5/2),(-2))

The equation of PQ 

use formula {{{((y-y1)/(y1-y2))=((x-x1)/(x1-x2))}}}

p=((11/2),(3/2)), Q=((5/2),(-2))

equation of PQ -> {{{(y-(3/2))/((3/2)-(-2))= (x-(11/2))/((11/2)-(5/2))}}}

{{{(y-3/2)/(7/2) = (x-(11/2))/3}}}

{{{3(y-(3/2))=(7/2)(x-(11/2))}}}

{{{3y-(9/2)=7x/2-(77/4)}}}

multiply equation by 4

{{{12y-18=14x-77}}}

14x-12y=59 is the equation of PQ