(8,5)(-12,-9)
slope = (-9 - 5) / (-12 - 8) = -14/-20 = 7/10
y = mx + b
slope(m) = 7/10
use either of ur points.....(8,5)....x = 8 and y = 5
now we sub and find b, the y int
5 = 7/10(8) + b
5 = 56/10 + b
5 - 56/10 = b
50/10 - 56/10 = b
-3/5 = b
so ur equation is : y = 7/10x - 3/5.....now we put it in standard form
y = 7/10x - 3/5...multiply by 10
10y = 7x - 6.....subtract 7x
-7x + 10y = -6....multiply by -1
7x - 10y = 6
So.....7x - 10y = 3.... IS NOT a good model. Because ur points are not a solution to ur equation.
Cos(A-B) = cosAcosB + sinAsinB
<span>
cos(</span>π/2 - θ) = cos(π/2)cosθ + sin(π/2)sinθ
π/2 = 90°
cos(π/2) = cos90° = 0. sin(π/2) = sin90° = 1
cos(π/2 - θ) = cos(π/2)cosθ + sin(π/2)sin<span>θ
</span>
= 0*cosθ + 1*sin<span>θ = </span>sin<span>θ
Therefore </span>cos(π/2 - θ) = sin<span>θ
QED </span>
