Question

the distance between the point (-4,3) and (-6,7) is​

Answer 1

Answer:

2√5 units

Step-by-step explanation:

x₁= -4, x₂ = -6, y₁= 3, y₂=7

Using the formula - √(x₂-x₁)²+(y₂-y₁)²

=> √[(-6-(-4)² + (7-3)²]

=> √(-2²+4²)

=> √20

= 2√5 units

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Answer 2

Coordinate Geometry

Clue: While solving this type of question it is needed to have the knowledge of one simple formula, which is as follows:

• The formula to calculate the distance between two points.

We've been provided with two points. With this information, we've been asked to calculate the distance between both points.

The given points are,

$\longrightarrow A(-4 3)$

$\longrightarrow B(-6, 7)$

So from here we can conclude that we have been given, $x_1 = -4\textsf{,} \; y_1 = 3 \; \; \& \; \; x_2 = -6\textsf{,} \; y_2 = 7.$

Solution:

Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane, then the distance between A and B is given by,

$AB = \sqrt{ {(x_{2} - x_{2}) }^{2} + {(y_{2} - y_{1})}^{2}}$

This is known as distance formula.

We can see that we have all the values which is needed in the formula. By using the distance formula and substituting the known values in it, we get;

$\implies AB = \sqrt{{(-6 - ( -4))}^{2} + {(7- 3)}^{2}}$

$\implies AB = \sqrt{{(-6 + 4)}^{2} + {(7- 3)}^{2}}$

$\implies AB = \sqrt{{(-2)}^{2} + {(4)}^{2}}$

$\implies AB = \sqrt{4 + 16}$

$\implies AB = \sqrt{20}$

$\implies \underline{ \underline{AB = 4.47}}$

Hence, the distance between of both points is 4.47 units.

$\rule{300}{2}$

Extra Information:

1. Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane, then the distance between A and B is given by,

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

2. The distance of the point P(x, y) from the origin O(0, 0) is given by,

$\longrightarrow OP = \sqrt{x^2 + y^2}$

3. Any point on the x-axis is of the form (x, 0).

4. Any point on the y-axis is of the form (0, y).