Find the point slope equation for the line that passes through the points (2,1) and (-2,5).

Mathematics

1 answer:

SVEN [57.7K]3 years ago

7 0

Answer:

y-1=-x+2

Step-by-step explanation:

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

m=(5-1)/(-2-2)

m=4/-4

m=-1

y-y1=m(x-x1)

y-1=-1(x-2)

y-1=-x+2

You might be interested in

How many feet are in 1 mile?

timurjin [86]

Answer:

5280, so not quite any of those, but the closest one is D

8 0

3 years ago

Read 2 more answers

Find the value of P:<br> -2+(-2)=1.2p+1/4p

Mrac [35]

1) -2 + (-2) = 1.2p + 1/4p
2) -4 = 1.2p + 0.25p
3) -4 = 1.45p
4) -4/1.45 = p

4 0

3 years ago

Given the following right triangle, find cose, sine, tane sececsce, and cote. Do not

IgorLugansk [536]

Step-by-step explanation:

Use the Pythagorean theorem:

$leg^2+leg^2=hypotenuse^2$

We have

$leg=4,\ hypotenuse=10$

Substitute:

$4^2+leg^2=10^2$

$16+leg^2=100$ <em>subtract 16 from both sides</em>

$leg^2=84\to leg=\sqrt{84}\\\\leg=\sqrt{(4)(21)}\\\\leg=\sqrt4\cdot\sqrt{21}\\\\leg=2\sqrt{21}$

$sine=\dfrac{opposite}{hypotenuse}\\\\cosine=\dfrac{adjacent}{hypotenuse}\\\\tangent=\dfrac{opposite}{adjacent}\\\\cotangent=\dfrac{adjacent}{opposite}\\\\secant=\dfrac{hypotenuse}{adjacent}\\\\cosecant=\dfrac{hypotenuse}{opposite}$

Substitute:

$hypotenuse=10,\ opposite=4,\ adjacent=2\sqrt{21}$

$\sin\theta=\dfrac{4}{10}=\dfrac{2}{5}\\\\\cos\theta=\dfrac{2\sqrt{21}}{10}=\dfrac{\sqrt{21}}{5}\\\\\tan\theta=\dfrac{4}{2\sqrt{21}}=\dfrac{2}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{2\sqrt{21}}{21}\\\\\cot\theta=\dfrac{2\sqrt{21}}{4}=\dfrac{\sqrt{21}}{2}\\\\\sec\theta=\dfrac{10}{2\sqrt{21}}=\dfrac{5}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{5\sqrt{21}}{21}\\\\\csc\theta=\dfrac{10}{4}=\dfrac{5}{2}$