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4 years ago

What’s the slope of the line that passes through (3,5) and (2,6)

2 answers:

8 0

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{6}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{6}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{2}-\underset{x_1}{3}}}\implies \cfrac{1}{-1}\implies -1$

Licemer1 [7]4 years ago

7 0

Answer:

Step-by-step explanation:

So what we need now are the two points you gave that the line passes through. Let's call the first point you gave, (3,5), point #1, so the x and y numbers given will be called x1 and y1. Or, x1=3 and y1=5.

Also, let's call the second point you gave, (2,6), point #2, so the x and y numbers here will be called x2 and y2. Or, x2=2 and y2=6.

Now, just plug the numbers into the formula for m above, like this:

6 - 5

2 - 3

or...

-1

or...

m=-1

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

y=-1x+b

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

(3,5). When x of the line is 3, y of the line must be 5.

(2,6). When x of the line is 2, y of the line must be 6.

Because you said the line passes through each one of these two points, right?

Now, look at our line's equation so far: y=-1x+b. b is what we want, the -1 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (3,5) and (2,6).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!.

You can use either (x,y) point you want..the answer will be the same:

(3,5). y=mx+b or 5=-1 × 3+b, or solving for b: b=5-(-1)(3). b=8.

(2,6). y=mx+b or 6=-1 × 2+b, or solving for b: b=6-(-1)(2). b=8.

See! In both cases we got the same value for b. And this completes our problem.

The equation of the line that passes through the points

(3,5) and (2,6)

y=-1x+8

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Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$