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3 years ago

Find the slope of the line that passes through the points (–4, 2) and (6, 6).

Mathematics

2 answers:

Studentka2010 [4]3 years ago

6 0

Answer:

Step-by-step explanation:

point P1 (-4,-2) in the form (x1,y1)

point P2(6,6) in the form (x2,y2)

slope = m

m = (y2-y1) / (x2-x1)

copy and paste the above formula into word pad on your computer for any problem asking about slope :)

m = [ 6-(-2) ] / [ 6-(-4) ]

m = [ 6+2 ] / [ 6+4 ]

m = 8 / 10

m = 4/5

Pavlova-9 [17]3 years ago

4 0

Answer:

2/5

Step-by-step explanation

rise over run:

$\frac{y_{2} - y_{2} }{x_{2} - x_{2}}$

$\frac{6-2}{6--4} = \frac{4}{10} = \frac{2}{5}$

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2 years ago

Find the area of the right triangle △DEF with the points D (0, 0), E (1, 1), and F.

Tems11 [23]

Answer:

The area of the triangle is $\sqrt{3}$

Step-by-step explanation:

Given:

Coordinates D (0, 0), E (1, 1)

Angle ∠DEF = 60°

△DEF is a Right triangle

To Find:

The area of the triangle

Solution:

The area of the triangle is = $\frac{1}{2}(base \times height)$

Here the base is Distance between D and E

calculation the distance using the distance formula, we get

DE = $\sqrt{(0-1)^2 + (0-1)^2}$

DE = $\sqrt{(-1) ^2 + (-1)^2$

DE = $\sqrt{1+1}$

DE = $\sqrt{2}$

Base = $\sqrt{2}$

Height is DF

DF = $tan(60^{\circ}) \times DE$

DF = $\sqrt{3} \times DE$

DF = $\sqrt{3} \times\sqrt{2}$

Now, the area of the triangle is

= $\frac{1}{2}({\sqrt{2})(\sqrt{3} \times \sqrt{2})$

= $\frac{1}{2}({\sqrt{2})(\sqrt{3} \sqrt{2})$

= $\frac{1}{2}(2\sqrt{3} )$

= $\sqrt{3}$

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2 years ago