✔ 最佳答案
(a) The mid-point of AC is (4,6). The slope of AC is (12-0)/(-2-10)=-1. So the slope of L is 1. Using point slope form, the equation of L is
y-6=x-4 or x-y+2=0
(b) The mid-point of BC is (3,-3). The slope of AM is (12+3)/(-2-3)=-3
So, using point slope form, the equation of AM is y-12=-3(x+2) or 3x+y-6=0
(c) Solve x-y+2=0...(1) and 3x+y-6=0...(2)
From (1) y=x+2. Sub. into (2) 3x+(x+2)-6=0
4x=4=>x=1 and y=3. P(1,3)
(d) Yes. Since the area of triangle AMB and the triangle AMC is the same (BM=MC with same height) Also the area of triangle PMB and the triangle PMC is the same (BM=MC with same height). So the area of (triangle AMB - triangle PMB) = area of (triangle AMC - triangle PMC) or The area of triangle APB = area of the triangle APC
To find out the area of triangle APB. First AP=√ [(1+2)^2+(3-12)^2]=√ 90. For the height, it is just the length of the vertical projection from B on the line AM, which is | 3(-4)+(-6)-6 /√ [1+(-3)^2] | =24/√ 10. So the area of triangle APB is [24/√ 10][√ 90]/2=36
