1題maths

a)
Since the gradient of AB (slope) =3
Using the point-slope form,
(y-6)/(x+1) = 3
y-6 = 3x+3
3x-y+9=0
Equation of AB is 3x-y+9=0.

For AC,
Slope of AC = (6-0)/(-1-2) = -2
Using the point-slope form,
(y-0)/(x-2) = -2
y = -2x+4
2x+y-4=0
Equation of AC is 2x+y -4=0

b)
Area of triangle =( Base x Height )/2
In this case, BC is the base
to find out the coordinates of B, put y=0 into the equation of AB, since B lies on the x-axis.
b = (-3,0)
Length of BC = 2-(-3) = 5
Height of the triangle = perpendicular distance between A and line BC
which is 6-0 = 6
Thus, Area of the triangle = (5 x 6 )/ 2 = 15 unit^2

2010-10-02 11:53:29 補充：
A kind reminder,
In exams, if the question asks you to find the area,
MAKE SURE that you write the unit of the area (in most cases, the unit is "unit square / unit^2 / square unit"
otherwise marks may be deducted.

2010-10-02 11:53:36 補充：
Similarly, if the question asks you to find the volume,
you should write the unit of the volume which is unit^3 / cubic unit.

I have made this mistake for many times.

1.
a) Find the equation of AB and AC. the equation of AB: (y-6)/(x+1)=3y-6=3x+3y=3x+9the equation of AC: (y-0)/(x-2)=(6-0)/(-1-2)y(-3)=6x-12y=-2x+4 b) Calculate the area of triangle ABCLet the coordinate at B be (x,0)(6-0)/(-1-x)=36/3=-1-x3=-xx=-3 the coordinate at B is (-3,0) the coordinate at C is (2,0) (given) BC=3+2=5 the height of ∆ABC is 6 (the coordinate of A is (-1,6)) the area of ∆ABC is 5*6/2=15