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3 years ago

P(5,-3) Q(2,4) find distance

Mathematics

1 answer:

Alenkinab [10]3 years ago

7 0

Answer: $\sqrt{58}$

The distance between the coordinates of the two points is measured by the Distance formula

Step-by-step explanation:

The distance between two points P(x1,y1) and Q(x2,y2) is measured by Distance formula i.e Distance= $\sqrt{}$ (x2-x1)²+(y2-y1)²

So, the distance between the two points P(5,-3) and Q(2,4) can be measured by

Distance PQ = $\sqrt{}$ (2-5)²+(4-(-3))²

PQ = $\sqrt{}$ (-3)²+7²

PQ= $\sqrt{}$ 9+49

PQ= $\sqrt{}$ 58

So,the distance of PQ is $\sqrt{58}$

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$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0$

Step-by-step explanation:

We want to write the trignometric expression:

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$\displaystyle \theta=\sec^{-1}\left(\frac{u}{10}\right)$

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$\displaystyle \sec(\theta)=\frac{u}{10}$

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$\displaystyle =2\left(\frac{\sqrt{u^2-100}}{u}\right)\left(\frac{10}{u}\right)$

Simplify. Therefore:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0$

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P(5,-3) Q(2,4) find distance​

P(5,-3) Q(2,4) find distance