Question 1196131
{{{ drawing ( 600, 600, -15, 15, -15, 15,
circle (-5,2,.12),locate (-5,2,A),
circle (2,3,.12),locate (2,3,B),
circle (-1,-6,.12),locate (-1,-6,C ),
circle (3,10,.12),locate (3,10,D ),

green(line(-5,2,3,10)),blue(line(2,3,-1,-6)),

graph( 600, 600, -15, 15, -15, 15, 0)) }}}


intersection of the AD and BC will be where their extensions intersect

so we need equations of the lines that containing AD and BC 

use endpoints of the line segment AD and find equation

slope is {{{m=(10-2)/(3-(-5))=8/8=1}}}

use point slope form to find equation


{{{y-y1=m(x-x1)}}}

plug in a slope and point D(-5, 2)

{{{y-2=1(x-(-5))}}}

{{{y-2=x+5}}}

{{{y=x+5+2}}}

{{{y=x+7}}} ... ...eq.1 

so we have an equation of a line that contains A and D


now find  a line that contains B and C


({{{2}}}, {{{3}}}), ( {{{-1}}}, {{{-6}}}) 


slope is {{{m=(-6-3)/(-1-2)=-9/-3=3}}}


use point slope form to find equation


{{{y-y1=m(x-x1)}}}

plug in a slope and point ({{{2}}}, {{{3}}})

{{{y-3=3(x-2)}}}

{{{y-3=3x-6}}}

{{{y=3x-6+3}}}

{{{y=3x-3}}}... ...eq.2

so we have an equation of a line that contains  B and C

the point M will be where these lines intersect


equal the right sides of the eq.1 and eq.2

{{{x+7=3x-3}}}  

{{{3+7=3x-x}}} 

 {{{10=2x}}} 

{{{x=5}}} 


now find {{{y}}}

{{{y=3*5-3}}}... ...eq.2

{{{y=12}}}

the point M will be at ({{{5}}}, {{{12}}})


let see it all on a graph


{{{ drawing ( 600, 600, -15, 15, -15, 15,
circle (-5,2,.12),locate (-5,2,A),
circle (2,3,.12),locate (2,3,B),
circle (-1,-6,.12),locate (-1,-6,C ),
circle (3,10,.12),locate (3,10,D ),
circle (5,12,.12),locate (5,12,M ),
green(line(-5,2,3,10)),blue(line(2,3,-1,-6)),

graph( 600, 600, -15, 15, -15, 15, 3x-3, x+7, 3x-3)) }}}