Question

Find the relation between x and y if A(x, y), B(-2.3) and C(2, 1) form an isosceles triangle with AB = AC.

Answer 1

Given: Points A(x, y), B(-2, 3) and C(2, 1), forming an isosceles triangle where AB = AC.

To find: A relation between x and y.

Answer:

We'll be using the distance formula for this.

Formula:

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

In this case, we're equating two sides. So the format will be:

$\tt \sqrt{\bigg(x_2\ -\ x_1\bigg)^2\ +\ (y_2\ -\ y_1\bigg)^2}\ =\ \sqrt{\bigg(x_3\ -\ x_2\bigg)^2\ +\ \bigg(y_3\ -\ y_2\bigg)^2}$

From the points given above, we have:

$\tt x_1\ =\ x\\\\x_2\ =\ -2\\\\x_3\ =\ 2\\\\y_1\ =\ y\\\\y_2\ =\ 3\\\\y_3\ =\ 1$

Using these in the formula,

$\tt \sqrt{\bigg(-2\ -\ x\bigg)^2\ +\ \bigg(3\ -\ y\bigg)^2}\ =\ \sqrt{\bigg(2\ +\ x\bigg)^2\ +\ \bigg(1\ -\ y\bigg)^2}\\\\\\\\\sqrt{\bigg(2\ +\ 2x\ +\ x^2\bigg)\ +\ \bigg(9\ -\ 6y\ +\ y^2\bigg)}\ =\ \sqrt{\bigg(4\ +\ 4x\ +\ x^2\bigg) +\ \bigg(1\ -\ 2y\ +\ y^2\bigg)}$

Squaring both sides,

$\tt 2\ +\ 2x\ +\ x^2\ +\ 9\ -\ 6y\ +\ y^2\ =\ 4\ +\ 4x\ +\ x^2\ +\ 1\ -\ 2y\ +\ y^2\\\\\\2\ +\ 2x\ +\ 9\ -\ 6y\ =\ 4\ +\ 4x\ +\ 1\ -\ 2y\\\\\\11\ +\ 2x\ -\ 6y\ =\ 5\ +\ 4x\ -\ 2y\\\\\\11\ -\ 5\ =\ 4x\ -\ 2x\ -\ 2y\ +\ 6y\\\\\\6\ =\ 2x\ +\ 4y\\\\\\\bf x\ +\ 2y\ =\ 3$