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2 years ago

What is the distance between (22,-3) and (1, -3)?

Mathematics

1 answer:

Bingel [31]2 years ago

6 0

Answer:

Option D

Step-by-step explanation:

<h2>Distance between two points</h2>

The y-value of the two points is unchanged. So, the line is parralel to x-axis.

The distance will be the diffrence between the x-co ordinates.

$\sf Distance = 1 - (-2\dfrac{1}{2})$

$\sf =1+\dfrac{5}{2}\\\\= \dfrac{2}{2}+\dfrac{5}{2}\\\\=\dfrac{7}{2}\\\\=3\dfrac{1}{2}$

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3 years ago

Use the slope formula to find the slope. y2 -4 y1 2 x2 0 x1 2 m=ΔyΔx=y2−y1x2−x1 Usetheslopeformula m = Δ y Δ x = y 2 - y 1 x 2 -

jenyasd209 [6]

Answer:

The slope is, 3

Step-by-step explanation:

Formula for slope(m) is given by:

$m = \frac{\triangle y}{\triangle x}$

$m = \frac{y_2-y_1}{x_2-x_1}$ .....[1]

Given the statement:

$y_2 = -4$

$y_1 = 2$

$x_2 = 0$ and $x_1 = 2$

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Simplify:

m = 3

Therefore, the value of slope is, 3

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3 years ago

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A line segment AB has the coordinates A (2,3) AND B ( 8,11) answer the following questions (1) What is the slope of AB? (2) What

GalinKa [24]

Answer:

(1) The slope of the line segment AB is 1. $\bar 3$

(2) The length of the line segment AB is 10

(3) The coordinates of the midpoint of AB is (5, 7)

(4) The slope of a line perpendicular to the line AB is-0.75

Step-by-step explanation:

The coordinates of the line segment AB are;

A(2, 3) and B(8, 11)

(1) The slope of a line segment is given by the following equation;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Where;

(x₁, y₁) and (x₂, y₂) are two points on the line segment

Therefore;

The slope, m, of the line segment AB is given as follows;

A(2, 3) = (x₁, y₁) and B(8, 11) = (x₂, y₂)

$Slope, \, m_{AB} =\dfrac{11-3}{8-2} = \dfrac{8}{6} = 1 \frac{1}{3} = 1.\bar3$

The slope of the line segment AB = 1. $\bar 3$

(2) The length, l, of the line segment AB is given by the following equation;

$l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}$

Therefore, we have;

$l_{AB} = \sqrt{\left (11-3 \right )^{2}+\left (8-2 \right )^{2}} = \sqrt{64 +36} = 10$

The length of the line segment AB is 10

(3) The coordinates of the midpoint of AB is given as follows;

$Midpoint, M = \left (\dfrac{x_1 + x_2}{2} , \ \dfrac{y_1 + y_2}{2} \right )$

Therefore;

$Midpoint, M_{AB} = \left (\dfrac{2 + 8}{2} , \ \dfrac{3 + 11}{2} \right ) = (5, \ 7)$

The coordinates of the midpoint of AB is (5, 7)

(4) The relationship between the slope, m₁, of a line AB perpendicular to another line DE with slope m₂, is given as follows;

$m_1 = -\dfrac{1}{m_2}$

Therefore, the slope, m₁, of the line perpendicular to the line AB, that has a slope m₂ = 4/3 = 1. $\bar 3$ is given as follows;

$m_1 = -\left (\dfrac{1}{\frac{4}{3} } \right ) = -\dfrac{3}{4} = -0.75$

The slope, m₁, of the line perpendicular to the line AB is m₁ = -0.75.

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3 years ago

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Answer:

Step-by-step explanation:

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