Question

Find the x value for point C such that AC and BC form a 2:3 ratio. Segment AB is shown. A is at negative 3, 5. B is at 3, 0. 6 −0.6 4 −2.4

Answer 1

the answer is -0.6 hope this help

Answer 2

Answer:

Hence, the x-value of point C is -0.6

Step-by-step explanation:

The coordinates of A are (-3,5)

and the coordinates of B are (3,0).

Point C cuts the line segment AB in the ratio 2:3.

If any point C cut the line segment AB with vertices A(a,b) and B(c,d) in the ratio m:n, then the coordinates of point C(e,f) is given by:

$e=\dfrac{mc+na}{m+n}$ and $f=\dfrac{md+nb}{m+n}$

Here we have m=2 and n=3

a=-3,b=5   and c=3 and d=0.

Hence, $e=\dfrac{2\times3+3\times(-3)}{2+3}=\dfrac{-3}{5}=-0.6$

and $f=\dfrac{2\times0+3\times5}{2+3}=\dfrac{15}{5}=3$

Hence the coordinates of point C is (-0.6,3).

Hence, the x-value of point C is -0.6.