Question 193036
): ACPB is a parallelogram. Given that A(-3,-2), B(0,9)and C(6,16). Find the co-ordinates of P and the area of CBAP

{{{drawing(400,400,-5,18,-5,18, locate(-.2,9.4,"@=B(0,9)"),
locate(-3-.2,-2+.4,"@=A(-3,-2)"), locate(6-.2,16.4,"@=C(6,16)"),
graph(400,400,-5,18,-5,18) 
 
)}}}

Connect these:

{{{drawing(400,400,-5,18,-5,18, locate(-.2,9.4,"@=B(0,9)"),
locate(-3-.2,-2+.4,"@=A(-3,-2)"), locate(6-.2,16.4,"@=C(6,16)"),
graph(400,400,-5,18,-5,18), line(-3,-2,0,9), line(0,9,6,16) )}}}  
 
We notice that C is 6 units right of B and 7 units above B

So the 4th vertex of the desired parallelogram will be 
6 units right of A and 7 units above A,

Since we want a point 6 units right of (-3,-2), we add 6 units 
to its x-coordinate -3 and get -3+6 or 3 for the x-coordinate
of P. And since we want the point to be 7 units above (-3,-2),
we add 7 units to its y-coordinate -2 and get -2+7 or 5 for the y-coordinate
of P.

So the fourth vertex of the parallelogram is P(3,5)

{{{drawing(400,400,-5,18,-5,18, locate(-.2,9.4,"@=B(0,9)"),
locate(-3-.2,-2+.4,"@=A(-3,-2)"), locate(6-.2,16.4,"@=C(6,16)"),
graph(400,400,-5,18,-5,18), line(-3,-2,0,9), line(0,9,6,16),
locate(3-.2,5.4,"@=P(3,5)")
 )}}}  

and the parallelogram looks like this:

{{{drawing(400,400,-5,18,-5,18, locate(-.2,9.4,"@=B(0,9)"),
locate(-3-.2,-2+.4,"@=A(-3,-2)"), locate(6-.2,16.4,"@=C(6,16)"),
graph(400,400,-5,18,-5,18), line(-3,-2,0,9), line(0,9,6,16),
locate(3-.2,5.4,"@=P(3,5)"), line(-3,-2,3,5),line(3,5,6,16)
 )}}}

To find the area of the parallelogram we first split it
into two congruent triangles using diagonal AC.

{{{drawing(400,400,-5,18,-5,18, locate(-.2,9.4,"@=B(0,9)"),
locate(-3-.2,-2+.4,"@=A(-3,-2)"), locate(6-.2,16.4,"@=C(6,16)"),
graph(400,400,-5,18,-5,18), 
locate(3-.2,5.4,"@=D(3,5)"), 
triangle(-3,-2,0,9,6,16),triangle(6,16,-3,-2,3,5)  )}}}

Then we find the area of triangle ABC by use of this determinant
formula:

{{{A = (1/2)abs(matrix(3,3,x[1],y[1],1,x[2],y[3],1,x[3],y[3],1)  )) }}}

where the points are taken counter-colckwise, that is A to C to B

{{{A = (1/2)abs(matrix(3,3,-3,-2,1,6,16,1,0,9,1)) =(1/2)45 = 22.5 ) }}}

So the area of the parallelogram is twice that or 45.

Edwin</pre>