Question

Find the equation of a locus of point whose distance from the point (2,-2) is equal to its distance from the line x-y=0

Answer 1

Find the equation of a locus of point whose distance from the point $(2, -2)$ is equal to its distance from the line $x - y = 0$

Solution:

The equation to a locus is the condition which the coordinates of each of the points of that locus and only those points must satisfy. In other words, the equation to a locus is nothing but the geometrical property expressed in algebraic language. Conversely, let $(x,y)$ represent any point on the locus, $x$ and $y$ satisfy the equation of the locus and every point satisfying the equation lies on the locus. If the functional relation between $x,y$ be $f(x,y) = 0$ , then we say $f(x,y) = 0$ is the Cartesian equation of the locus. Thus in coordinate geometry a locus is represented by an equation. The set of points $(x,y)$ satisfying a given equation $f(x,y) = 0$ is called the locus or graph of the equation.

In our case we have to find the equation for locus of a point $(2, -2)$ is equal to its distance from the line $x - y = 0$ (or we can write $x = y$ )

Let $P(x, y)$ be a point satisfying the geometrical condition, call point $A(2, -2)$ and $O(0, 0)$ .

We can graph the line $x - y = 0$

$PA = OP$

$$\begin{array}{l} = > PA^{2} = OP^{2} \end{array}$$

So we can find from equation of distance between points:

$$\begin{array}{l} (x - 2)^{2} + (y + 2)^{2} = x^{2} + y^{2} \\ x^{2} - 4x + 4 + y^{2} + 4y + 4 = x^{2} + y^{2} \\ -4x + 4 + 4y + 4 = 0 \\ 4y + 8 = 4x \\ y = x - 2 \end{array}$$

Answer: $y = x - 2$