Question

What is the distance between points (11, 3) and (4, 3) on a coordinate plane?

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Answer 1

Solution:

To solve this problem, we need to know Distance Formula.

Distance Formula:

Let P(x₁, y₁) and Q(x₂, y₂) be two points on the Cartesian Plane. Then the distance between the two points is given as:-

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

Here, the points are: (11, 3) and (4, 3)

Therefore, the distance between the points will be:

$\rm\longrightarrow Distance = \sqrt{(11-4)^{2}+(3-3)^{2}}$

$\rm\longrightarrow Distance = \sqrt{(7)^{2}+(0)^{2}}$

$\rm\longrightarrow Distance = \sqrt{49}$

$\rm\longrightarrow Distance =7\ unit.$

→ Therefore, the distance between the points (11, 3) and (4, 3) is 7 unit.

Answer:

• Distance between the points (11, 3) and (4, 3) is 7 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:

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

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:

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

3. Centroid of a triangle.

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:

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