Question

find distance a point (9,12)formt he origin

Answer 1

Solution:

To solve this problem, we have to know the distance formula.

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane. Then the distance between P and Q is given by the formula:

$\tt\longrightarrow D =\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Given coordinate is: (9, 12)

Coordinate of origin: (0, 0)

Plugging the values into the formula, we get:

$\tt\longrightarrow D =\sqrt{(9-0)^{2}+(12-0)^{2}}$

$\tt\longrightarrow D =\sqrt{9^{2}+12^{2}}$

$\tt\longrightarrow D =\sqrt{81+144}$

$\tt\longrightarrow D =\sqrt{225}$

$\tt\longrightarrow D =\sqrt{(15)^{2}}$

$\tt\longrightarrow D=15\:\: unit.$

So, the distance of the point (9, 12) from the origin is 15 unit.

Answer:

• The distance of the point (9, 12) from the origin is 15 unit.

Learn More:

1. Section formula.

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the point which divides PQ internally in the ratio m₁ : m₂. Then, the coordinates of R will be:

$\tt\longrightarrow R = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)$

2. Mid-point formula.

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R will be:

$\tt\longrightarrow R = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)$

3. Centroid of a triangle formula.

Centroid of a triangle is the point where the medians of the triangle meet.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle. Let R(x, y) be the centroid of the triangle. Then, the coordinates of R will be:

$\tt\longrightarrow R = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)$