Question

Triangle ABC below is reflected and dilated such that the coordinates of B' are (-2,-1).



What will be the coordinates of A'?

Answer 1

Since B' is reflected and dilated from B, we can find the original coordinates of B by reversing the dilation and reflection process.

Let's assume that the dilation factor is k, and the reflection is across the line y = x. This means that the coordinates of B are (-2/k, -1/k), where (-2, -1) is the reflected and dilated image of B.

To find the coordinates of A, we need to apply the same dilation and reflection process to A'.

Let the coordinates of A' be (x, y).

First, we need to find the relationship between the coordinates of A and B. Since A and B are corresponding points in the original and reflected/dilated triangles, we can write the following proportion:

(AB') / (AB) = (A'B') / (A'B)

Using the distance formula to calculate the lengths of the sides:

√[(x - (-2))^2 + (y - (-1))^2] / √[(x - a)^2 + (y - b)^2] = √[(x - (-2/k))^2 + (y - (-1/k))^2] / √[(x - a/k)^2 + (y - b/k)^2]

Simplifying this equation:

√[(x + 2)^2 + (y + 1)^2] / √[(x - a)^2 + (y - b)^2] = √[(x + 2/k)^2 + (y + 1/k)^2] / √[(x - a/k)^2 + (y - b/k)^2]

Squaring both sides:

[(x + 2)^2 + (y + 1)^2] / [(x - a)^2 + (y - b)^2] = [(x + 2/k)^2 + (y + 1/k)^2] / [(x - a/k)^2 + (y - b/k)^2]

Cross-multiplying:

[(x + 2)^2 + (y + 1)^2] * [(x - a/k)^2 + (y - b/k)^2] = [(x + 2/k)^2 + (y + 1/k)^2] * [(x - a)^2 + (y - b)^2]

Expanding and simplifying both sides:

(x^2 + 4x + 4 + y^2 + 2y + 1) * ((x - a/k)^2 + (y - b/k)^2) = (x^2 + 4x + 4/k^2 + y^2 + 2y + 1/k^2) * (k^2 * (x - a)^2 + k^2 * (y - b)^2)

(x^2 + 4x + 4 + y^2 + 2y + 1) * (x^2 - 2ax/k + a^2/k^2 + y^2 - 2by/k + b^2/k^2) = (x^2 + 4x + 4/k^2 + y^2 + 2y + 1/k^2) * (k^2 * x^2 - 2ak^2 * x + a^2 + k^2 * y^2 - 2bk^2 * y + b^2)

Expanding further:

x^4 + x^3(4 - 2a/k) + x^2(4 + a^2/k^2 + 4/k^2) + x(4 - 2a/k) + 4/k^2 + 4a^2/k^2 + x^2(2 - 2b/k) + xy(4 - 2b/k) + y^2(1 + 1/k^2) + y(2 - 2b/k) = k^2 * x^4 + x^3(-2ak^2) + ax^2 + k^2 * y^4 + y^3(-2bk^2) + by^2 + k^2 * (2 - 2a)xy

Now, we can see that for the corresponding sides of the equation, the coefficients of x^4, x^3, x^2, x, y^4, y^3, y^2, y, and xy must be equal.

Comparing coefficients:

1 = k^2
4 - 2a/k = -2ak^2
4 + a^2/k^2 + 4/k^2 = ax^2
4 - 2a/k = k^2 * (2 - 2a)
4/k^2 + 4a^2/k^2 = k^2
2 - 2b/k = 0
2 - 2b/k = 0
1 + 1/k^2 = 1

From these equations, we can solve for k and a to find their values. Once we know the values of k and a, we can substitute them back into the equations to solve for x and y, which will give us the coordinates of A'.