Question

George is making a triangular enclosure in his yard. He plans on using two old fence posts as vertices, and will place a new

Answer 1

Answer:

The correct answers for the possible locations for the are;

(-9, 1)

(0, 0)

Step-by-step explanation:

The coordinates of two of the three posts are given in feet as (-5, 4) and (2, 6)

The length of the available fencing = 25 feet

The length, l, of the segment between the coordinates of the two old posts vertices of the fence is given by the following equation;

$l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$

Where;

(x₁, y₁) = (-5, 4)

(x₂, y₂) = (2, 6)

$l = \sqrt{\left (6-4 \right )^{2}+\left (2-(-5) \right )^{2}} = \sqrt{53} \approx 7.25 \ feet$

Given the coordinates of the third point as (x₃, y₃), we have;

Therefore, we have;

$\sqrt{\left (y_{3}-4 \right )^{2}+\left (x_{3}-(-5) \right )^{2}} + \sqrt{\left (y_{3}-6 \right )^{2}+\left (x_{3}-(2) \right )^{2}} = 25 - \sqrt{53}$

For the point (2, 6), we have;

$\sqrt{\left ((-6)-4 \right )^{2}+\left (2-(-5) \right )^{2}} + \sqrt{\left ((-6)-6 \right )^{2}+\left (2-(2) \right )^{2}} = 24.2 > 25-\sqrt{53}$

For the point (-9, 1), we have;

$\sqrt{\left ((-6)-4 \right )^{2}+\left (2-(-5) \right )^{2}} + \sqrt{\left ((-6)-6 \right )^{2}+\left (2-(2) \right )^{2}} = 17.08 < 25-\sqrt{53}$

For the point (0, 0), we have;

$\sqrt{\left ((0)-4 \right )^{2}+\left (0-(-5) \right )^{2}} + \sqrt{\left ((0)-6 \right )^{2}+\left (0-(2) \right )^{2}} = 12.73 < 25-\sqrt{53}$

For the point (8, 8), we have;

$\sqrt{\left ((8)-4 \right )^{2}+\left (8-(-5) \right )^{2}} + \sqrt{\left ((8)-6 \right )^{2}+\left (8-(2) \right )^{2}} = 19.92 > 25-\sqrt{53}$

For the point (-4, -5), we have;

$\sqrt{\left ((-5)-4 \right )^{2}+\left ((-4)-(-5) \right )^{2}} + \sqrt{\left ((-5)-6 \right )^{2}+\left ((-4)-(2) \right )^{2}} = 22.04 > 25-\sqrt{53}$

Therefore, the correct answers for the possible locations for the are (-9, 1) and (0, 0).