Question

If the distance from A (5,6) to B (1, b) is twice the distance from B to

Answer 1

Answer:

The possible values of $b$ are -2.944 and -9.055, respectively.

Step-by-step explanation:

From statement we know that $AB = 2\cdot BC$. By Analytical Geometry, we use the equation of a line segment, which is an application of the Pythagorean Theorem:

$AB = 2\cdot BC$

$\sqrt{(x_{B}-x_{A})^{2}+(y_{B}-y_{A})^{2}} = 2\cdot \sqrt{(x_{C}-x_{B})^{2}+(y_{C}-y_{B})^{2}}$ (1)

Where:

$x_{A}$, $x_{B}$, $x_{C}$ - x-Coordinates of points A, B and C.

$y_{A}, y_{B}, y_{C}$ - y-Coordinates of points A, B and C.

$(x_{B}-x_{A})^{2}+(y_{B}-y_{A})^{2} = 4\cdot (x_{C}-x_{B})^{2}+4\cdot (y_{C}-y_{B})^{2}$

Then, we expand and simplify the expression above:

$x_{B}^{2}-2\cdot x_{A}\cdot x_{B} +x_{A}^{2} +y_{B}^{2}-2\cdot y_{A}\cdot y_{B} + y_{A}^{2} = 4\cdot (x_{C}^{2}-2\cdot x_{C}\cdot x_{B}+x_{B}^{2})+4\cdot (y_{C}^{2}-2\cdot y_{C}\cdot y_{B}+y_{B}^{2})$

$x_{B}^{2}-2\cdot x_{A}\cdot x_{B} + x_{A}^{2} +y_{B}^{2}-2\cdot y_{A}\cdot y_{B} + y_{A}^{2} = 4\cdot x_{A}^{2}-8\cdot x_{C}\cdot x_{B}+4\cdot x_{B}^{2}+4\cdot y_{C}^{2}-8\cdot y_{C}\cdot y_{B}+4\cdot y_{B}^{2}$

If we know that $x_{A} = 5$, $y_{A} = 6$, $x_{B} = 1$, $y_{B} = b$, $x_{C} = 1$ and $y_{C} = -3$, then we have the following expression:

$1 -10 +25 +b^{2} -12\cdot b+36 = 100 -8 +4 +36+24\cdot b +4\cdot b^{2}$

$b^{2}-12\cdot b +52 = 4\cdot b^{2}+24\cdot b +132$

$3\cdot b^{2}+36\cdot b +80 = 0$

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:

$b_{1} \approx -2.944$, $b_{2} \approx -9.055$

In consequence, the possible values of $b$ are -2.944 and -9.055, respectively.