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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.

Feb 10, 2022

#1
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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?

Feb 10, 2022
#2
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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?

Guest Feb 10, 2022
#3
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wait a second, that would be incorrect considering the x for S should be smaller than 8

Guest Feb 10, 2022
#4
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haha, sorry about the mistake from before, it was a pretty careless mistake, i forgot you can just subtract 5 from 8 inorder to get the X coordinate. So in this case, S would be (3,0)?

Guest Feb 10, 2022
#5
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so if S= (3.0) and the distance between the two is 5, if you add 5 to 0 you will get 5 for the y coordinate of R, (considering the fact the line is vertical). So the final answer would be 5-3, which is 2?

Guest Feb 10, 2022