Question

Find the distance between the points (0, 4) and (-4,0).​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

Given Points: (0, 4) and (-4, 0)

Here:

$\rm:\longmapsto \begin{cases}\rm x_{1} = 0\\ \rm x_{2} = - 4\\ \rm y_{1} =4 \\ \rm y_{2} = 0\end{cases}$

Distance between two points is calculated by using the formula given below:

$\rm:\longmapsto D=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$

Substitute the value in the formula, we get:

$\rm:\longmapsto D=\sqrt{(4)^{2}+(4)^{2}}$

$\rm:\longmapsto D=\sqrt{2 \times (4)^{2}}$

$\rm:\longmapsto D=4\sqrt{2} \: unit.$

★ Therefore, the distance between the two points (0, 4) and (-4, 0) is 4√2 units.

$\textsf{\large{\underline{Answer}:}}$

• The distance between the two points (0, 4) and (-4, 0) is 4√2 units.

$\textsf{\large{\underline{Learn More}:}}$

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:

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

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:

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

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:

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

Answer 2

Answer:

HELLO MATE,

Step-by-step explanation:

$[tex]\textsf{\large{\underline{Solution}:}}$

Given Points: (0, 4) and (-4, 0)

Here:

$\rm:\longmapsto \begin{cases}\rm x_{1} = 0\\ \rm x_{2} = - 4\\ \rm y_{1} =4 \\ \rm y_{2} = 0\end{cases}$

Distance between two points is calculated by using the formula given below:

$\rm:\longmapsto D=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$

Substitute the value in the formula, we get:

$\rm:\longmapsto D=\sqrt{(4)^{2}+(4)^{2}}$

$\rm:\longmapsto D=\sqrt{2 \times (4)^{2}}$

$\rm:\longmapsto D=4\sqrt{2} \: unit.$

★ Therefore, the distance between the two points (0, 4) and (-4, 0) is 4√2 units.

$\textsf{\large{\underline{Answer}:}}$

The distance between the two points (0, 4) and (-4, 0) is 4√2 units.

$\textsf{\large{\underline{Learn More}:}}$

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:

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

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:

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

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:

$\rm:\longmapsto R = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)$[/tex]