Answer:
The possible values of [tex]b[/tex] are -2.944 and -9.055, respectively.
Step-by-step explanation:
From statement we know that [tex]AB = 2\cdot BC[/tex]. By Analytical Geometry, we use the equation of a line segment, which is an application of the Pythagorean Theorem:
[tex]AB = 2\cdot BC[/tex]
[tex]\sqrt{(x_{B}-x_{A})^{2}+(y_{B}-y_{A})^{2}} = 2\cdot \sqrt{(x_{C}-x_{B})^{2}+(y_{C}-y_{B})^{2}}[/tex] (1)
Where:
[tex]x_{A}[/tex], [tex]x_{B}[/tex], [tex]x_{C}[/tex] - x-Coordinates of points A, B and C.
[tex]y_{A}, y_{B}, y_{C}[/tex] - y-Coordinates of points A, B and C.
[tex](x_{B}-x_{A})^{2}+(y_{B}-y_{A})^{2} = 4\cdot (x_{C}-x_{B})^{2}+4\cdot (y_{C}-y_{B})^{2}[/tex]
Then, we expand and simplify the expression above:
[tex]x_{B}^{2}-2\cdot x_{A}\cdot x_{B} +x_{A}^{2} +y_{B}^{2}-2\cdot y_{A}\cdot y_{B} + y_{A}^{2} = 4\cdot (x_{C}^{2}-2\cdot x_{C}\cdot x_{B}+x_{B}^{2})+4\cdot (y_{C}^{2}-2\cdot y_{C}\cdot y_{B}+y_{B}^{2})[/tex]
[tex]x_{B}^{2}-2\cdot x_{A}\cdot x_{B} + x_{A}^{2} +y_{B}^{2}-2\cdot y_{A}\cdot y_{B} + y_{A}^{2} = 4\cdot x_{A}^{2}-8\cdot x_{C}\cdot x_{B}+4\cdot x_{B}^{2}+4\cdot y_{C}^{2}-8\cdot y_{C}\cdot y_{B}+4\cdot y_{B}^{2}[/tex]
If we know that [tex]x_{A} = 5[/tex], [tex]y_{A} = 6[/tex], [tex]x_{B} = 1[/tex], [tex]y_{B} = b[/tex], [tex]x_{C} = 1[/tex] and [tex]y_{C} = -3[/tex], then we have the following expression:
[tex]1 -10 +25 +b^{2} -12\cdot b+36 = 100 -8 +4 +36+24\cdot b +4\cdot b^{2}[/tex]
[tex]b^{2}-12\cdot b +52 = 4\cdot b^{2}+24\cdot b +132[/tex]
[tex]3\cdot b^{2}+36\cdot b +80 = 0[/tex]
This is a second order polynomial, which means the existence of two possible real solutions. By Quadratic Formula, we have the following y-coordinates for point B:
[tex]b_{1} \approx -2.944[/tex], [tex]b_{2} \approx -9.055[/tex]
In consequence, the possible values of [tex]b[/tex] are -2.944 and -9.055, respectively.
