Question

example 5: a) show that the right bisectors of the sides of δdef all intersect at po of the triangle.
b) verify that point c is equidistant from the three vertices of δdef.
(chart with triangle def and point c(-4,4), d(-18,12), e(-6,-12), f(12,6) on a coordinate grid)

Explanation:

Part b) Verify that point \( C(-4, 4) \) is equidistant from the three vertices \( D(-18, 12) \), \( E(-6, -12) \), and \( F(12, 6) \) of \( \triangle DEF \).

To verify equidistance, we use the distance formula: For two points \( (x_1, y_1) \) and \( (x_2, y_2) \), the distance \( d \) between them is:

$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Step 1: Distance from \( C(-4, 4) \) to \( D(-18, 12) \)

Let \( (x_1, y_1) = (-4, 4) \) and \( (x_2, y_2) = (-18, 12) \).
Substitute into the distance formula:

$$ CD = \sqrt{(-18 - (-4))^2 + (12 - 4)^2} = \sqrt{(-14)^2 + (8)^2} = \sqrt{196 + 64} = \sqrt{260} $$

Step 2: Distance from \( C(-4, 4) \) to \( E(-6, -12) \)

Let \( (x_1, y_1) = (-4, 4) \) and \( (x_2, y_2) = (-6, -12) \).
Substitute into the distance formula:

$$ CE = \sqrt{(-6 - (-4))^2 + (-12 - 4)^2} = \sqrt{(-2)^2 + (-16)^2} = \sqrt{4 + 256} = \sqrt{260} $$

Step 3: Distance from \( C(-4, 4) \) to \( F(12, 6) \)

Let \( (x_1, y_1) = (-4, 4) \) and \( (x_2, y_2) = (12, 6) \).
Substitute into the distance formula:

$$ CF = \sqrt{(12 - (-4))^2 + (6 - 4)^2} = \sqrt{(16)^2 + (2)^2} = \sqrt{256 + 4} = \sqrt{260} $$

Since \( CD = CE = CF = \sqrt{260} \), point \( C \) is equidistant from all three vertices of \( \triangle DEF \).

Answer:

Point \( C(-4, 4) \) is equidistant from \( D \), \( E \), and \( F \) because \( CD = CE = CF = \sqrt{260} \).