Triangle ABC has vertices A(0,8), B(2,0), C(8,0). A vertical line intersects AC at R and {BC} at S, forming triangle RSC. If the area of RSC is 12.5, determine the positive difference of the x and y coordinates of point R.
+23245 Let O be the origin.
Triangle(OAC) is similar to triangle(SRC).
OA = 8 and OC = 8 making OA = OC.
Because of similar triangles, SR = SC; call this length "x".
Area(triangle(SRC)) = ½·x·x = 12.5 ---> x2 = 25 ---> x = 5
The distance from S to R is 5.
The distance from C to S is also 5.
Can you take it from here?
Yes!
so we can use the distance formula which is √ [ (x₂ - x₁)² + (y₂ - y₁)²], and plug S's coordinate in along with C, we get
S= (13,0) correct?
wait a second, that would be incorrect considering the x for S should be smaller than 8
