Analytical Geometry

something is missing in 1.1 1nd 1.2, in paticular the coordinates of point D.

In 1.1 you can determine the equation of the line (y=mx+b) however m or b will remain unknown (it takes two points to determine a line)

In 1.2 you can show that (3/2, -3/2) is equidistant from points A and C, but this is only a single point among a family of points (interestingly this family is also a line). Moreover, if you use (3/2, -3/2) as point D, this describes line AD.

1.3 two points are known, so you can describe line AC easily by using y=mx+b and solving for m and b.

1.4 you now have lines AC and BD as y=m1x+b1 and y=m2x+b2 and if m1 is not equal to m2 then they cross at some point. this point is determined by substitution, graphically, elimination, or matrix methods.

1.5 once lines AC and AD are determined the angle CAD can be determined.

1.6 several methods can be used to determine triangle area. If it were me, I'd determine the length of the legs and use Heron's formulae.

Rap

The co ords of point D are meant to be missing, you must work them out. Actually they give them to you in Q 1.2 (Stupid)

Solution though: (I think)

1.1) Gradiet of BC = y2 - y1/x2 - x1
= 3 -(-1)/2-6
= -1
(Thats kind of obvious just by looking at the line however)
Since AD is parallel to BC, m of AD is also 1.

Therefor the equation od AD: y = mx +c
y = -1x + c
(-3;3): 3 = -1(-3) + c
0 = c

Therefor equation of AD is: y = -x


1.2) DB² = DC² (Equidistant)
(x-2)² + (y -3)² = (x-6)² + (y +1)²
x² -4x + 4 + y² -6y +9 = x² -12x +36 + y² +2y +1
-4x -6y +13 = x² -12x +y² +2y + 37
-8y = -8x +24 (Principles of completing the square)
y = x - 3



D is pat od the line AD and AD's ofrmula is y = -x

Therfore: -x = x -3
-2x = -3
x = 3/2

So then y = -(3/2)

So your co ords for D are (3/2 ; -3/2)

Ok 1.3 was easy. Get the gradient of AC which is 3 -(-1)/-3-6 = 4/-9

So the Eq is: y = mx +c
y = 4/-9 x +c
I used A's co ords for substitution so:
(-3;3): 3 = (4/-9)(-3) +c
3 = + 4/3 + c
X3: 9 = +4 +3c
5 = 3c
5/3 = c

So Eq is y = 4/-9x + 5/3


Now it's 1.4 and 1.5 which I just can't get. Hopefully I'm not using incorrect values for them which I'v calculated above. They all seem feasible though. Sorry but I don't know Heron's formula either :/

Oh Yeah

if A is (3,-3) and D is (-3/2,3/2) then AD has to go through the origin (0,0) and the slope is (-1). Then AD is y=-x.

The rest of then then just falls out. However, the attached drawing isn't so obvious, cause it doesn't show AD going through the or1gin.

Not that I'm complaining, mind you (see I too can be obtuse).

Rap

And the condition that AD is parallel to BC is missing from the problem statement. Now if AB were parallel to DC then ABCD would be a rhombus and the area would be the product of the length of the diagonals. A fact that was known to Euclid, 2000 years before DeCarte and analytic geometry (more obstrafugation).

Rap

1.4 Intersection of AC & BD

first develop eqns of lines (y=mx+b) between AC & BD

AC (3,-3) & (6,-1) m=-4/9 & b=5/3 or y=-4/9x+5/3
a little rearrangement this becomes 4x+9y=15

BD (2,3) & (3/2,-3/2) m=9 & b-15 or y=9x-15
a little rearrangement this becomes 9x-y=15

Now there are two linear equations in two unknowns and by substitution (I used matrix elimination*) I can determine that these lines cross at (30/17,15/17)

matrix elimination
|4 9||x|=|15|
|9 -1||y|=|15|

|4 9||x|=| 15|
|81 -9||y|=|135|

|85 0||x|=|150|
|81 -9||y|=|135|

or x=150/85=30/17
then y=9(30/17)-15=15(18/17-1)=15/17 by substitution.

1.5 takes geometry and trig

BCD is an isosceles triangle and the legs DB & CD are equal length. So a triangle from the midpoint of BC and vertex D form two right triangles out of BCD, and from trig you know that cos(theta)=1/2 the length of BC/length of DB (or CD). so theta is arcos(BC/(2DB))

you can determine these two lengths from the point distance formulae d=[(x1-x2)^2+(y1-y2)^2]^(1/2)

Rap

Thanks for you help there. My meathod was way off.