Question

Show that point A is equidistant from the endpoints of segment BC

Answer 1

To demonstrate that point A is equidistant from the endpoints of segment BC, we need to prove that the distances from point A to both endpoints, B and C, are equal. Mathematically, this can be expressed as |AB| = |AC|, where |AB| represents the distance between points A and B, and |AC| represents the distance between points A and C.

Step-by-step explanation:

Firstly, let's denote the coordinates of points B, C, and A in a Cartesian coordinate system as (xB, yB), (xC, yC), and (xA, yA) respectively. The distance between two points (x1, y1) and (x2, y2) in a Cartesian plane is given by the distance formula:

`\[ \text{Distance} = √((x2 - x1)^2 + (y2 - y1)^2) \]`

Now, we want to prove that |AB| = |AC|. Using the distance formula, we can express this as:

`\[ √((xB - xA)^2 + (yB - yA)^2) = √((xC - xA)^2 + (yC - yA)^2) \]`

Squaring both sides of the equation eliminates the square root:

`\[ (xB - xA)^2 + (yB - yA)^2 = (xC - xA)^2 + (yC - yA)^2 \]`

Expanding and simplifying both sides should lead to an expression that demonstrates the equality of the distances. If the two sides are equal, it confirms that point A is equidistant from the endpoints of segment BC.

Answer 2

for this case we can solve the question with the following operations:

`AB=\sqrt[]{(0-8)^2+(-7-7)^2}=\sqrt[]{260}`

And similar:

`AC=\sqrt[]{(-8-8)^2+(-5+7)^2}=\sqrt[]{260}`

Then since AB = AC we can conclude that A is equidistant from the endpoints of the segment BC