In right triangle $ABC$, $M$ and $N$ are midpoints of legs $\overline{AB}$ and $\overline{BC}$, respectively. Leg $\overline{AB}$ is 6 units long, and leg $\overline{BC}$ is 8 units long. How many square units are in the area of $\triangle APC$?
+128408 Let B =(0,0) A = (0,6) and C = (8,0) M =(0,3) N = (4,0)
Let the slope of MC = [ 3 -0 ] / [ 0 - 8] = -3/8
And the equation of MC is y = -(3/8) x + 3 (1)
Let the slpoe of AN = [ 6 - 0 ] / [ 0 - 4 ] = -6/4 = -3/2
And the equation of An is y = (-3/2)x + 6 (2)
Set (1) = (2) to find the x coordinate of P
(-3/8)x + 3 = (-3/2)x + 6
(-3/8 + 3/2)x = 3
(9/8)x = 3
24/9 = x = 8/3
So...the y coordinate of P is (-3/2)(8/3) + 6 = -24/6 + 6= -4 + 6 = 2 = altitude of triangle PNC
The area of triangle APC = area of triangle ANC - area of triangle PNC
Area of ANC =(1/2) (AB) (NC) = (1/2)(6)(4) = 12 units^2
Area of PNC = (1/2) (2)(NC) = (1/2) (2) (4) =4 units^2
So...area of triangle APC = 12 - 4 = 8 units^2
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