Question

point a lies at (2,4) and point b lies (3,6) which point needed so that teiangle abc is isosceles

Answer 1

To create an isosceles triangle ABC with points A(2, 4) and B(3, 6), point C must be positioned such that at least two sides of the triangle are equal in length.

1. **Calculate the length of AB:**

The distance between points A and B can be calculated using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Using A(2, 4) and B(3, 6):
\[
AB = \sqrt{(3 - 2)^2 + (6 - 4)^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}
\]

2. **Choose a point C such that AC = BC:**

Let's denote C as (x, y). We want to find the coordinates of point C such that AC = BC.

- The distance AC:
\[
AC = \sqrt{(x - 2)^2 + (y - 4)^2}
\]

- The distance BC:
\[
BC = \sqrt{(x - 3)^2 + (y - 6)^2}
\]

For the triangle to be isosceles, we need:
\[
\sqrt{(x - 2)^2 + (y - 4)^2} = \sqrt{(x - 3)^2 + (y - 6)^2}
\]

Squaring both sides:
\[
(x - 2)^2 + (y - 4)^2 = (x - 3)^2 + (y - 6)^2
\]

Expanding both sides:
\[
(x^2 - 4x + 4) + (y^2 - 8y + 16) = (x^2 - 6x + 9) + (y^2 - 12y + 36)
\]

Simplifying:
\[
x^2 - 4x + 4 + y^2 - 8y + 16 = x^2 - 6x + 9 + y^2 - 12y + 36
\]

Canceling \(x^2\) and \(y^2\) from both sides:
\[
-4x + 20 - 8y = -6x + 45 - 12y
\]

Rearranging gives:
\[
2x - 4y = 25 \quad \text{(1)}
\]

From this equation, we can express y in terms of x:
\[
4y = 2x - 25 \implies y = \frac{2x - 25}{4} \quad \text{(2)}
\]

3. **Choose a specific value for x to find corresponding y**:

Let's say we choose \(x = 5\):
\[
y = \frac{2(5) - 25}{4} = \frac{10 - 25}{4} = \frac{-15}{4} = -3.75
\]

Thus, one point C that makes triangle ABC isosceles is \(C(5, -3.75)\).

You can also choose different values for \(x\) to get other points for C. The general form for point C, derived from equation (2), will yield various coordinates by plugging in different \(x\) values.