Question

Distance between the points P(x, -y) and Q(-x, 2y) is

Answer 1

Answer:

PQ = $\sqrt{4x^{2} +9y^{2} }$

Explanation:

First of all, let us have a look at the distance formula to find the distance between 2 points which have coordinates as $(x_1,y_1)$ and $(x_2,y_2)$.

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

Here we have two points:

$P(x, -y)$ and $Q(-x, 2y)$

Comparing with the coordinates $(x_1,y_1)$ and $(x_2,y_2)$:

$x_2 = -x\\x_1 = x\\y_2 = 2y\\y_1 = -y$

Now, let us apply the above formula to find the distance between the points P(x, -y) and Q(-x, 2y).

Putting the values in distance formula to find the distance PQ:

$PQ = \sqrt{(-x-x)^2+(2y-(-y))^2}\\\Rightarrow PQ = \sqrt{(-2x)^2+(2y+y)^2}\\\Rightarrow PQ = \sqrt{(-2x)^2+(3y)^2}\\\Rightarrow PQ = \sqrt{4x^2+9y^2}$

So, the answer is:

Distance between the points P(x, -y) and Q(-x, 2y) is: PQ = $\sqrt{4x^{2} +9y^{2} }$

Answer 2

$Distance\ between\ P\ and\ Q=\sqrt{4x^2+9y^2}$

Explanation:

Given:

Points

P(x, -y)

Q(-x, 2y)

Find:

Distance between P and Q

Computation:

Distance between two points $=\sqrt{(x2-x1)^2+(y2-y1)^2}$

$Distance\ between\ P\ and\ Q=\sqrt{(-x-x)^2+(2y+y)^2} \\\\Distance\ between\ P\ and\ Q=\sqrt{(-2x)^2+(3y)^2}\\\\Distance\ between\ P\ and\ Q=\sqrt{4x^2+9y^2}$