To form an isosceles triangle with points A (2, 4) and B (3, 6), we need to find a point C such that the distances AC and BC are equal.
Let's calculate the distance between A and B first:
- \( A(2, 4) \)
- \( B(3, 6) \)
Using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(3 - 2)^2 + (6 - 4)^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]
Now, we will check each of the given points (C) to see if it creates an isosceles triangle (where AC = BC).
-
C (3, 4):
- Distance AC: \[ AC = \sqrt{(3 - 2)^2 + (4 - 4)^2} = \sqrt{1 + 0} = 1 \]
- Distance BC: \[ BC = \sqrt{(3 - 3)^2 + (6 - 4)^2} = \sqrt{0 + 4} = 2 \]
(AC ≠ BC)
-
C (0, 4):
- Distance AC: \[ AC = \sqrt{(0 - 2)^2 + (4 - 4)^2} = \sqrt{4 + 0} = 2 \]
- Distance BC: \[ BC = \sqrt{(0 - 3)^2 + (4 - 6)^2} = \sqrt{9 + 4} = \sqrt{13} \]
(AC ≠ BC)
-
C (5, 4):
- Distance AC: \[ AC = \sqrt{(5 - 2)^2 + (4 - 4)^2} = \sqrt{9 + 0} = 3 \]
- Distance BC: \[ BC = \sqrt{(5 - 3)^2 + (4 - 6)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \approx 2.83 \]
(AC ≠ BC)
-
C (5, 5):
- Distance AC: \[ AC = \sqrt{(5 - 2)^2 + (5 - 4)^2} = \sqrt{9 + 1} = \sqrt{10} \]
- Distance BC: \[ BC = \sqrt{(5 - 3)^2 + (5 - 6)^2} = \sqrt{4 + 1} = \sqrt{5} \]
(AC ≠ BC)
None of the points create an isosceles triangle with the distances calculated.
Let’s analyze if there can be other possibilities. For the triangle to be isosceles with AC = BC, the point C must lie on the perpendicular bisector of segment AB.
The midpoint M of AB is:
\[ M = \left(\frac{2 + 3}{2}, \frac{4 + 6}{2}\right) = \left(\frac{5}{2}, 5\right) = (2.5, 5) \]
The slope of line AB is:
\[ \text{slope} = \frac{6 - 4}{3 - 2} = \frac{2}{1} = 2 \]
The slope of the perpendicular line is the negative reciprocal:
\[ \text{slope of perpendicular} = -\frac{1}{2} \]
The equation of the line (perpendicular bisector) through M (2.5, 5) is:
Using the point-slope form \(y - y_1 = m(x - x_1)\):
\[ y - 5 = -\frac{1}{2}(x - 2.5) \ y - 5 = -\frac{1}{2}x + 1.25 \ y = -\frac{1}{2}x + 6.25 \]
Now we can check if any of the point choices fit this line, but since the points we checked didn't yield equality, we conclude:
No point A (3, 4), B (0, 4), C (5, 4), or D (5, 5) creates an isosceles triangle with A (2, 4) and B (3, 6).
If strictly restricted to the answer choices given, none satisfies the condition outlined for an isosceles triangle, but one solution could theoretically exist based on the perpendicular bisector derived from the midpoints. To conclusively determine the characteristics of the triangle, further exploration may be required.